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THE
Elements of Algebra
BY
GEORGE LILLEY, Ph.D., LL.D.
EXPRESIDENT SOUTH DAKOTA AGRICULTURAL COLLBOB
OF THE ^
UNIVERSITY
SILVER, BURDETT & COMPANY
New York . . . BOSTON . . . Chicago
1894
Copyright, 1892,
By Silver, Burdett and Company.
mniijcrsitg lirrss :
John Wilson and Son, Cambridqk, U.S.A.
PREFACE.
Algebra is a means to be used in other mathematical
work; it develops the mathematical language, and is the
great mathematical instrument. If the student would be
come a mathematician, he must understand this language
and possess facility in handling the various forms of literal
expressions.
Attention is called to the sequence of subjects as herein
presented. Involution is introduced as an application of
multiplication, evolution as an application of division, and
logarithms as an application of exponents. Throughout the
book the student is led to see that one subject follows as an
application of another subject. The beginner is led to see at
the outset that Algebra, like Arithmetic, treats of numbers.
Algebraic terms and definitions are not introduced until
the student is required to put them into actual use. Correct
processes are cleariy set forth by carefully prepared solutions,
the study of which leads the pupil to discover that method
and theory follow directly from practice, and that methods
are merely clear, definite, linguistic descriptions of correct
processes.
The book is sufficiently advanced for the best High Schools
and Academies, and covers sufficient ground for admission to
any American College.
Great care has been given to the selection and arrangement
of numerous examples and problems. These have been, for
18:5963
iv PREFACE.
the most part, tested in the recitationroom, and are not so
difficult as to discourage the beginner.
It remains for the author to express his sincere thanks
to W. H. Hatch, Superintendent of Schools, Moline, 111. ; to
Professor W. C. Bojden, SubMaster of the Boston Normal
School, Boston, Mass. ; and to O. S. Cook, connected with
the literary department of Messrs. Silver, Burdett & Co., for
reading the manuscript and for valuable suggestions.
GEORGE LILLEY.
Pullman, Washington, June, 1892.
PREFACE TO THE SECOND EDITION.
In this edition the typographical errors have been cor
rected, and a page of examples has been added to Chapter
XXVII ; also, the exercises have been carefully revised and
corrected. Answers to the examples and problems have been
prepared, and are bound in the book, or separately in flexi
ble cloth covers. The answerbook is furnished for the use
of the class only on application of teachers to the publishers
for it. The publishers and the author desire to express their
appreciation of the very favorable reception accorded to the
first edition.
September, 1894.
CONTENTS.
CHArrEK PACK
I. First Principles l
II. Algebraic Addition 19
III. Algebraic Subtraction 27
IV. Algebraic Multiplication 35
V. Involution 52
VI. Algebraic Division 60
VII. Evolution 79
VIII. Use of Algebraic Symbols 99
IX. Simple Equations 104
X. Problems Leading to Simple Equations 109
XI. Factoring 119
XII. Highest Common Factor 141
XIII. Lowest Common Multiple 155
XIV. Algebraic Fractions '164
XV. Fractional Equations •. . . 201
XVI. Simultaneous Simple Equations 215
XVII. Problems Leading to Simultaneous Equations . . 238
XVIIL Exponents 248
XIX. Radical Expressions 263
XX. Logarithms 296
XXI. Quadratic Equations 312
vi CONTENTS.
CHAPTEK PAGE
XXII. Equations which may be Solved as Quadratics . 330
Theory of Quadratic Equations 339
XXIII. Simultaneous Quadratic Equations 345
XXIV. Indeterminate Equations 355
XXV, Inequalities 363
XXVI. Series 373
Arithmetical. 373
Geometrical 379
Harmonical 384
XXVII. Ratio and Proportion 388
APPENDIX 401
INDEX TO DEFINITIONS.
PAOI
Algebra 118
Binomial 46
Coefficient 20
Equation, Biquadratic . 334
" Degree of, Roots of 35, 108
" Exponential 309
Literal 206
" Syminetriciil 347
Expression, Algebraic 90
" Compound 23
** Homogeneous 349
** Imaginary 286
** Irrational 263
** Mixed 164
" Simple 21
Factor 119
Figures, Subscript 227
Fraction, Complex 188
Continued 190
Identities 104
Index 79
Mean, Arithmetical 378
" Geometrical 383
" Harmonical 385
Nlonoraial 21
Multiple ,155
" Common 155
Multiplication, Algebraic 38
Vlll INDEX TO DEFINITIONS.
PAGE
Numbers, Algebraic, Absolute ... 14
" Known 107
" Negative 11
" Scale of 12
" Unknown 107
Polynomial 23
Power 35
Progression, Arithmetical 373
" Geometrical 379
" Harmonical 384
Quantity <.....• 389
Keciprocal ^8
Roots 79, 340
Signs, Algebraic 13
" Double 80
" Law of . 38, 61
" Radical 79
Subtraction ' 34
Surd, Similar 263
" Entire, Mixed 264
" Quadratic 290
Symbols of Abbreviation 7
" of Aggregation, of Relation 6
" of Operation 1, 99
Terms 3, 90
" Like 20
Term, Absolute 207
" Degree of. Dimension of 345
Value, Absolute • < 14
** Numerical 9
ELEMENTS OF ALGEBRA
CHAPTER I.
FIRST PRINCIPLES.
1. In Algebra figures and letters are used to represent
numbers, instead of figures, as in Arithmetic.
Thus, we may use x to represent the number of dollars in a man's
business, the number of cents in the cost of an article, the number of
miles from one place to another, the number of persons in our class,
etc.
In Algebra, the letter x is reasoned about and operated upon just
the same aa the numbers which it represents are reasoned about and
operated upon in Arithmetic.
2. Symbols of Operation. The signs +, — , X, and h,
are used to deuute the algebraic operations addition, sulv
traction, multiplication, and division, that in Arithmetic
can actually be performed. + is read 'plus; — is read
miniis ; X is read multiplied by; H is read divided by.
A dot or point is sometimes used instead of the sign X.
Thus, a X 6 and a • h both mean that a is to be multiplied
by h. The multiplicand is usually written before the
multiplier.
Dimsioii in Algebra is more frequently represented by
placing the dividend as the numerator, and the divisor as
2 ELEMENTS OF ALGEBRA.
the denominator of a fraction. Thus, a i b, or  , means
b
that a is to be divided by b. Eead a divided by b.
m
Note. Do not read such expressions as — , m over n; it is meaningless.
3. We must be careful to distinguish between arith
metical and algebraic operations. The former can actually
be performed, whereas many operations in Algebra can only
be indicated.
Thus, suppose a man owes ^ 5 for a vest and % 20 for a coat, actual
addition gives $25 as his total indebtedness. But if the number of
dollars he owes for the vest be represented by m, and the number of
dollars that he owes for the coat be represented by n, his entire debt
can only be indicated. In order to show that the number represented
by m is to be added to the number represented by n, we use the sign
+ written between them ; thus, m + n.
Exercise I.
Eead the following algebraic expressions :
1. 0! + 100 ; a + 10  2 ; &  2 ; &  100 + 8.
2. a \b; m \ n { ^ \ m + s ~ r; a — b + m.
3. c + 2x5; c 10 X2; snX r20.
4. q + t + S X m; c + m^n — s  q; —  + c^a—p { 1  x.
a
Indicate by means of algebraic expressions the following:
5. The sum of m and n. The difference between m
and n. The sum of x, y, and a.
6. The sum of m, n, and r diminished by t. If you had
m cents, earned n cents, and are given r cents, and then
spend t cents ; how many cents will you have left ?
FIRST FRINCIPLKS. 3
7. John has m apples, Henry has n apples, and Charles
has b apples ; express the number of their apples. How
many more have John and Henry than Charles ?
8. If you buy goods for a dollars and sell them at a gain
of b dollars, express the selling price.
9. I buy goods for m dollars and sell them at a loss of
71 dollars ; express my selling price.
10. Henry had x marbles ; he gave John vi marbles, and
Charles ii marbles. How many had he left ?
11. I pay n cents for a reader, x cents for a history, y
cents for a grammar, 6 cents for carfare, and have m cents
left ; express the number of cents that I had at first.
12. A boy earned a dollars, then received m dollars
from his father, n dollars from his mother ; and spent k
dollars of what he had for books, x dollars for a coat, and
y dollars for a sled. Express the number of dollars he had
left.
4. The Sign of Multiplication is generally omitted in
Algebra, except between figures. Thus,
bah means bXaXb, prstuz means p X r X s X t X v X z ;
J • 3 • 4 • 5 means 2 X 3 X 4 X 5, or 120.
Again, if the numher of gallons in a cisk of cider is represented by
a, and the number of cents in the cost of one gallon is represented
by m, then the number of cents in the cost of n casks is represented
by amn.
5. In the expression 5 + 2«o — a+ — — : 5,
m 2 am ^
2 ab, a, — , and ttt &re called Terms.
n ooc
ELEMENTS OF ALGEBRA.
Exercise 2.
Eead and state the meaning of the following algebraic
expressions :
1. 5ahx\ ah. Result : 5 times a times h times
c
X, plus 7n times n divided by c, minus a times b ; etc.
2. kl + 1t; PQrs + ab cd + mnx y — 80.
en b
, ^ ^ ^klx abed a. i . ii
3. amnpgr — cdXo\ ; Zb d+ 11— r.
, bwyz mnop
4 bx'^+12pqrstQ^hk^a\imz.
ab
„ S ab d — 10 mnr + Imnr st
5. .
ad — 1
6. h + '^^+u; 2^+2^^+^^. ?±fZ^* + „ + *_±f .
4 y u X a I
6. It is customary to write the letters in the order of
the alphabet.
In a product represented by several letters and numbers, the num
bers are written first. Thus,
cX&XaX5X3 is written 3 X 5 ahc ; both mean 15 a 6 c.
Also, s X r X n X m X 25 is written 25 mnr s.
Exercise 3.
Write algebraic expressions for the following :
1. The product of x, y, and z ; of m, n, and 5 ; of 3 and
xy, of 5, a, b, and S X mn,
2. The product of a and b divided by their sum. Their
product divided by their difference.
FIRST PHINCIPLES. 5
3. The product of m, n, r, and 25 divided by the sura of
m and n. The same product divided by the difference of
VI and 71.
4. A travels at the rate of 3 miles an hour ; how many
hours will it take him to travel 30 miles ? How many
hours to travel a miles ? To travel m n miles ? To travel
60 a in n miles ?
5. A man bought 18 loads of wheat, of m bushels each,
at n cents a bushel ; how many cents in the entire cost ?
6. In example 5, suppose that he sold the wheat at a
gain of r cents a bushel ; how many cents did he gain ?
How many cents in the selling price ?
7. In example 5, suppose that he sold the wheat at a
loss of a cents a bushel ; how many cents would he lose ?
how many cents in the selling price ?
8. A man bought a boxes of peaches, each containing h
peaches, at c cents a peach ; and m baskets of grapes, each
containing n pounds, at r cents a pound. How many cents
did he pay for both ?
9. A man worked n hours a day for m days, at a cents
an hour. With the money he bought a coat for x cents ;
how many cents had he left?
10. One boy sold a apples at c cents each ; another sold
n peaches at m cents each ; a third sold r peais at t cents
each. How many cents did they all receive ?
11. I buy 5 tons of coal at SIO per ton, and pay for
it in cloth at S2 per yard ; how many yards will it take?
I buy a tons of coal at h dollars per ton, and pay for it in
cloth at m dollars a yard ; how many yards will it take ?
6 ELEMENTS OF ALGEBRA.
12. A man works n weeks at h dollars a week, and his
son works m weeks at r dollars a week. With the money
they pay for c cords of wood at d dollars a cord ; how many
dollars have they left ?
13. If 5 cords of wood cost $15, how many dollars will
3 cords cost? If c cords cost $m, how many dollars will
n cords cost ?
14. A man drove 3 hours at the rate of 10 miles an
hour ; how many hours will it take him to walk back at
the rate of 6 miles an hour ? If he drives 3 days n hours
each day, at the rate of t miles an liour, and 5 days m hours
each day, at the rate of s miles an hour, how many hours
will it take him to return over the same distance, at the
rate of r miles an hour ?
15. If you buy t tons of coal at the rate oi %d for n
tons, and sell it at a loss of $ Z on each ton, how many
dollars will you receive ? Suppose you sell at a gain of
%h on each ton, how many dollars will you get for it ?
Suppose you sell all of it for r dollars, and make a profit,
how many dollars profit will you get ?
7. Symbols of Relation. The signs =, >, and <, are
used for the words, equals, is greater than, and is less than,
respectively.
Symbols of Aggregation. The signs ( ) , [ ] , { } , and ,
are used to show that the terms enclosed by them are to
be treated ns one number. They are called parenthesis,
bracket, hracc, and vinculum, respectively. Thus,
(2 ft + 6) (3 X  y), [2 a + h][3x  yl {2 a + b] {3 x ~ i/],
2a i b X 3x — y, each shows that the number obtained })y adding
the terms 2 a and b is to be multiplied by the result obtained by
subtracting y from 3 x.
FIRST PRINCIPLES. 7
Sjrmbols of Abbreviation. The signs (of deduction) .*.,
(of reason) •.*, and (of continuation) ...., are used for the
words, heiice or therefore, since or hecmise, and so on,
respectively.
8. Since 81 = 9 X 9, or written 9^ for brevity, 81 is
called the second power of 9. Since 27 = 3 X 3 X 3, or
written 3^ for brevity, 27 is called the third power of 3.
Similarly a^, {m n)\ (m + n)^, are called second poioers of
a, m n, and m ^ n\ also a^, (m n)^, {m \ n)^, are third
powers of a, m n, and m \ n. cfi means a X a ; a^ means
a X a X a \ etc. In general, ft" is called the nth power of
a, read a 7ith power.
9. In the expression a^ + h^c^ — 3 x"; 2, 4, 5, and ?i
are called Exponents, h^ c^ means bxbxbxbxcxc
xcX cXc; 4 and 5 are used for convenience to show how
many times b and c are used as factors.
We must be careful to keep in mind the meaning of each indicated
operation when rea(hng an algebraic expression. Thus, the expres
sion 5 X* j/"^ — 2 a* 6 («' — 6*)^ + 3a^c*d"* means, five times the third
power of X times, the second power of y^ minus two tim^s the fourth
power of a times b times the fifth power of tlie expression in the
parenthesis, a seventh power minus b sixth power, phis three times
the fifth power of a times the fourth power of c times the with power
of d.
Exercise 4.
Read the following :
1. m^;3m^x^;5m^nY,'fh(M^(^iab^hh,vL^n^\ lOaVr^.
2. m^n^ f 5 a^bxi/  3 m^^ x^; m^n^ 'lab m n + (H'K
3. 10 (ft 6)^0; (?n3n8) (m nf\ (a^  „)2. (,„2 _ 3 ^^^a
4. (m n  m3)3; 3 (v^ b {<i  }?'^ (n^ + />") (a^  b^f
8 ELEMENTS OF ALGEBRA.
5. 3 oTlf', 3 {a bf ; a^ (b^  c^  d'^f^ (m^ n^) (m n)\
6. (10 m + n"^) (10 n^  m^f < 15 a (x  i/)^ (x + yf ;
(1^ + 1)^ + d^ + ^7 (c5 + d^y
b{a + h + ef ^ ^'' ^ c2 + 6^3 '
7. •.•a + 2:c=:6 + ^, .'.x = h — a\ {a'^ — c''f={7n?+n^f,
... a"*  C'* =r m2 ^ ^2 . ^2  ^3 = 2 ^3 _ 2 ,^,2^ ..0.^3 = ,^,^2 .
x\x^x\'X\ to n terms =^ nx\ a X a X a X a X
.... to 71 factoi'S = a" ; 1 ^ x ■\ x^ + st? \ ....
..2 I ^, ^^ I f^ ., f^,„«o _ ^('''" ~ ^)
1 a?
a + ar + a ?'^ + a ?^ + . . . . to 7«. terms = , .
r — 1
Write algebraic expressions for the following :
8. The sum of m and n. The double of x. The second
power of the sum of a and h. The second power of differ
ence between x and y. Five times the third power of the
difference of x and y.
9. The second power of the sum of x second power
and y. The second power of the sum of x and y second
power. The product of the fourth power of x, the third
power of ?/, and the second power of 7n. The product of
the first power of x and three times the nth power of y.
The product of x second power plus y second power, and
X second power minus y second power.
10. The product of the sum of x second power and y
by n a. Five n third power minus seven m n plus six a
second power, m third power minus two times b second
power c plus n fourth power is equal to n times y.
11. Seven times m fourth power times n second power
minus two times ^t seventh power times m third power
plus three times a tliird power times h second power plus
eight times a second power times b third power plus five
FIRST PRINCIPLES. 9
times a fifth power. Since a plus h equals m minus w,
therefore the second power of a plus h is equal to the
second power of m minus ii.
12. Therefore, x is equal to m third power, because x
plus three m third power is equal to two x plus two m
third power, a plus a plus a, and so on to n minus two
terms, equals n minus two times a. The second power
of m plus yi, divided by m minus n is less or greater than
m times a plus h plus c plus o? plus e. a less than 6 is
equal to iii greater than n.
13. A horse eats a bushels and an ox h bushels of oats
in a week ; how many bushels will they together eat in n
weeks ? If a man was a years old 50 years ago, how old
will he be x years hence ?
10. The Nomerical Value of an algebraic expression is
the number of positive or negative units it contains, and is
found by giving a particular value to each letter, and then
peiforming the operations indicated. Thus,
If a  3, 6 = 4, x = 5, y = 6, find the numerical values of :
,2,. 9 6x«
25 a* 1/2
Replacing? the letters in each expression hy the •particular values
given for them, we have
Process. 4 a«6« = 4 x 3^ X 4»
= 4X9X64
= 2304.
9h3* 9 X 4 X 5«
25rt»y2~25 X 3»X 62
9 X 4 X 125
25 X 27 X 36
= A
27'
U. If one factor of a product is equal to 0, the whole
product nmst be equal to 0, wlmtever values the other factors
10 ELEMENTS OF ALGEBRA. ,
may have ; and it is also clear that no product can be zero
unless one of the factors is zero. Thus, ah is zero if a is
zero, or if b is zero; and if ab is zero, either a or 6 is zero.
Again, if a; = 0, then a^h'^xy^ = 0, also ax{y^ + 62; + «^)
== 0, whatever be the values of a, 7;, 3/, and z.
Exercise 5.
If <x = 6, & = 2, c = 1, a; = 5, ?/ — 4, find the numerical
values of the following algebraic expressions :
L 3c2; 72/3; 5a&; 9^?/, 8Z?3. 3^5. ^^8.7^4. ^^^.10. 3 ^^4^
2. 9 6*; 2 a a; ; 3/^ , 10 x^ ; \y^\ 5 6?/; ^ r^^ ; ^^ab cxy.
3. 3^2 ^^ 5 1^ ^ ^ ; 7 c^ ; f .«3 ; a* x^ ; 8 a^^ 2 3/* ;  acxy.
li 7n = 2, n = 3, p = I, q = 0, r = 4, s = 6, find the
values of:
4 ^^. Pi,„,2^. 4^^^■^ ?ZL!!_^ ^J!^. o«»9n. 2m2g
p; 8^'^^ r, 6 06 5 s 4 5 7776 r 27 m' 64 2'
9 7?t3 ' 6 ' ' /^ ' 54 ^m ' 32 ' r» ' 3"
Example 6. Find the value of 5 6^ + j% x y  5 a^  ^a^b^,
when (1 = 2, 6 = 3, x = 5, and i/ = 10.
Replacing the letters in the expression by the particular values
given for them, we have
Process.
563 4 ^3_xy _ 5a2_ I Q^'2^3 = 5 X 33 +^3_ X 5 X 10  5 X 22  I X 22 X 33
= 5X27 + 3X5 5X43x27
= 49.
Example 7. Find the value of mny^\ mhi xyt + m"nh's — j^ ,
wlien ?7i = 5, u = 2, r = 3, s = 4, X = 0, and y = I.
FIRST PRINCIPLES. 11
Process.
mny*Hm*nxytfm%V8^ = 5 X 2 X IHO + S'X 2» X 3<X4 ^ ^ ^.^
25 X«
= 5X2X1 + 25X8X81X4——^
= 10 4 64800  50
= 64760.
If a = 1, b = 2, c = 3, and d = 0, find the numerical
values of the following algebraic expressions :
8. 10 a — 4:b { 6 c \ 5 d] ab + hc + ac — da,
9. 6 abScd{ 10 a d2b c+ 2bd; 2 be + 10 cd.
10. a^ + Ir^\c^(P; abc + 10 bed + 5 ac d + 'S abd.
.11. a* + h^ \ c'd; +ob8e { ad; —40ad\ab.
12. 5a+3cG6+6(/; 36co?2acc?10rt&rf.
13. 5/>r3 + .^3 ^_ /,3_ i.25a&3c; 15 «2 ^_ jj. r* + 10 a &.
.Cw/2/,4 .IftS
14. c3 8 «(/6^ 5 ftio^. j^^y.
c3 ai« 62 c '
15. 125a6cc/*m + ^,^^l^'; «» 4. ?,3 + ,3 + ,^;^.
80 X
V\ 2 ., ?,3 4. iL ^ 2;{c  ^^, 3 «2i3,;6^ ^ '_' _ 2^'
/; a'
8 , , 3
) a
** 2^'^ c^ abc
8 ^/ /•" 8<* ^'^ r^
19. J a f^ ;/ \ I n —  ^^2 d^xy\ Wa^^^^ '^ ~q~ ~ ^
o o
12. Negative Numbers, if a person owes a debt of ten dollars,
and \\di» but .six dollars in money, he can pay the debt only in part.
12
ELEMENTS OF ALGEBRA.
For his six dollars in money will cancel only six dollars of his debt,
and leave him still owing four dollars; we may consider him as be
ing worth four dollars less than nothing. The total number of dol
lars that he is worth may be represented by — 4, because it will take
four dollars in addition to the six dollars to pay the debt. If a person
gains eight dollars and loses eleven dollars, the number of dollars in
his net loss may be represented by — 3, because it will take three dol
lars in addition to the gain to balance the loss. Similarly, if he gains
100 dollars and loses 120 dollars, the number of dollars in his net loss
may be represented by — 20. To enable us to represent these num
bers, it is necessary to assume a new series of numbers, beginning at
zero and descending in value from zero by the repetitions of the unit,
precisely as the natural series ascends from zero. To each of these
numbers the sign — is prefixed. The negative series of numbers is
written thus :
10, 9, 8, 7,
6,
1, 0.
For convenience the algebraic series of numbers is represented as
follows :
Scale of Numbers. We may conceive algebraic numbers
as measuring distances from a fixed point on a straight line,
extending indefinitely in both directions, the distances to
the right being positive, and the distances to the left nega
tive. From any point on the line, measuring tovjard the
right is positive and tovmrd the left negative.
^h
lis
ilo
l5
llO
+ 115
In the above illustration consider A the zero or startingpoint on
the scale of numbers, and the distance between any two consecutive
numbers one unit. The distances to the right and left of A are posi
^
i
FIRST PHINCIPLES. 13
tive (+) and negative (— ), respectively, as indicated by the direc
tions of the arrows.
To add + 9 to + 4 (read 9 and 4 in tlie positive series) ^ we start at
4 in the positive series, count nine units in the positive direction, and
arrive at 13 in the positive series. That is, + 4 + (+ 9) = 13.
To add + 9 to — 4 (iead 9 in the positive series and 4 in the negative
series) f we begin at 4 in the Jiegative series^ count nine units in the
positive directioiiy and arrive at 5 in the positive series. That is,
4f (+9) = +5.
To add — 9 to I 4, we start at 4 in the positive series, count nine
units in the negative direction, and arrive at 5 in the negative series.
That is, + 4 + ( 9) =  5^
To add — 9 to — 4, we start at 4 in the negative series, count nvie
units toward the left, and arrive at 13 in the negative series. That
is,  4 + ( 9) =  13.
To subtract + 9 from + 4, we start at 4 in the positive series, count
nine units in the negative direction, and arrive at 5 in the negative
series. That is, + 4  (+ 9) =  5.
To subtract + 9 from — 4, we begin at 4 in the negative series,
count nine units in the negative direction, and arrive at 13 in the
negative series. That is — 4 — (4 9) = — 13.
To subtract — 9 from + 4, we begin at 4 in the positive series,
count nine units in the positive direction, and arrive at 13 in the posi
tive series. That is, + 4  ( 9) = + 13.
To subtract —9 from —4, we start at 4 in the negative series, count
nine units in the positive direction, and arrive at 5 in the positive
series. That is,  4  ( 9) =  5.
13. The sign + is often omitted before a number in the positive
series. Thus, the numbers 3, 5, and 6, taken alone, mean the same
^ (+3), (f 5), and (+ 6), showing that the numbers are in the
positive series
The sign — must always be written when a number is in the
negative series. Thu.s, the numbers 3, 5, and 6, taken in the negative
series, are written ( 3), (  5), and ( 6).
The Algebraic Signs f and — mark the direction that
the numbers following them are to take. These signs are
14 ELEMENTS OF ALGEBRA.
used to indicate opposition (opposite direction), also opera
tion. The former is called the positive, and the latter the
negative sign.
An Algebraic Number is one Which is represented by an
algebraic term vnth its sign of direction. Thus, + 3,3,
— a, and + 5 a are algebraic numbers.
Absolute Value shows what place a number has in the
positive or negative series. Thus, + 3 and — 3 have the
same absolute value ; that is, three units.
Absolute Numbers are those not affected by the signs f
or —.
Example. The meaning of an algebraic expression, as
3a;2+(2a6)[c(2/)],
is explained thus :
To 3 x^ units in the positive series add 2ab units in the negative
series, and from their sum subtract the expression in brackets, c in the
positive series minus y in the negative series. The signs written
before the terms (—2 a 6), (— y), and before the bracket, indicate
operation. The sign written before 2 ah and y, also the sign under
stood before 3 x^ and c, indicate opposition.
Exercise 6.
1. Over how many units and in what series of numbers
would a point move in passing from 43 to — 8? — 10 to
+ 1? +5 to +15? 12tol? lto12? 15 to 5?
9 to 9 ? — 5 to — 5 ?
2. Which is the greater, or — 6 ? 3 or — 3 ? — 5 or
— 3 ? + 10 or — 1 ? + 50 or — 50, and how many units ?
3. How many units is + 6 greater than 0, + 3, — 3, — 6,
and — 5 ? How many units is — 5 less than 5 ? How
many units is a less than b ?
FIRST PRINCIPLES. 15
4 If a point start at (f 3) and move three units to the
right, then five units to the left, where is it? Express its
distance from 0.
5. Suppose a point start at (+ 2), and move six units
to the right, then eleven units to the left, where is it ?
6. Where is the point which, starting at (— 5), moved
(— 3), then (+ 8) ? Express its distance from the starting
point.
7. Suppose a point starting at + 3, move + 2, then — 7,
then f 5, then — 6, then + 10, then — 11, where would it
be ? Express its distance from I 3.
Explain the meaning of:
8. 2 [3 6 f ( 5 a)]  5 [( a) + (+ h)].
9. (+ 8a;) 4 [+ 3x(+ 12 y) + ( x)  (8 y)].
Also the meaning of the signs + (as used or understood)
and — .
Explain the meaning of:
10. 6«5A2 f (\a^}fi) f (^aH^c^) + (aH'^) + (+aH^)
+ 20 a2^c2 + ( aH^) + ( a^b^).
11. aU^ ^ ^_ ^^.10) ^ (^ ^6 ^5^ ^_ (_ y a.) + ,;i3 ,^ft
 (4 a^° m^).
12. + (+ a^) ~{\h^) (+ a^)  (+ ^).
13. (x + 7/)2 + (a + xf  (x + 7/)2  (a + x)^
Find their numerical values when a^I?^c^m = n = k
= y = x=l.
16 ELEMENTS OF ALGEBRA.
Eead the following expressions, and find their numeri
cal values when a = 0, b = 1, c = 2, d = S, e = 4,
n = 6, and m = 6 :
14. c5  (+c)  (+ n^) (+ c)  (+ d^) + a^(+ h^) (+ <^\
15. 3 [g + (+ 7ij\  5 (+ c) h (a + 6) + 2 (+ «) ^ 6.
16. (+c) [a + (4 ?i) + (+^)  (+6)]  (+ m) r (+ d),
17. [(+m)  (+rO + (+m)  (+e)(+c)]  [(+m) (+6)].
18. d^^{^d')^'2{+h')^{+c'){+c^)^de^^{\4:%
If tt = 5, & = 4, c = 3, t^ = 2, and e = 1, find the numeri
cal values of the following expressions :
19. (+a2)4(+/^3)_(+c2)(+e5); a'^cahl?+d^^{a'^W').
If a = 1, 5 == 2, c = 3, c? = 4, c = 5, find the values of:
26+2 3c9 e2l d^ 8a24.3?>2 4,2^552 c^^.^
e3 62"^e + 3' F' «2+Zy2 + c262 ^2
d>h'' eh c ' b'^+d^bd' e^+ed+d^'
a^ + 4a%+6a%'^ + 4:ab^{ h\ 28 12
a^ + Sa^b + Sal)^ + b^ ' o? + h^j\c^^ d?c^}p^'
24. aJ— 156^5; — ^ ^ ; .
a \ b c + ft o + c
a*4a3(?46«2c2_4r?.^3__^4 ,^ , ,^ ^, ^^
25. ,4 ^xq . ...22 .z.q 7 4 ; 12e4a^(2aX5)26.
b*—4:Wc+bb^c^—4:b(f\c^ ^ ^
26. [(12e4a)^2«] XZ^; a^ ^ {aHU^) ^ (^c^  a'^y
29.
FIRST PRINCIPLES. 17
27. Ti ;, , > + ^ . . , , ; (« + ^) (6 + c).
28. (6 + c) (c + rf) + (c + rf) (rf + e) + ^_^ji^_^.
30. 66^(a — c) — 3rf + a6ctZi24a;c + 5f]^Ho
+ a X e.
Express the following statements in algebraic symbols :
31. To the double of a add 6.
32. To five times x add h diminished by one.
33. Increase h by the sum of a and x divided by y.
34. Write x, a times.
What is the sum oi x\ x^x ■\ .... written a times ?
35. Write three consecutive numbers of which n is the
least.
36. Write five consecutive numbers of which m is the
greatest
37. Write w, a minus 1 times ; also m plus n times.
38. Write seven consecutive numbers of which x is the
middle one.
39. Write a, icth power, minus y, nth power.
40. To the double of x, increased by a divided by &, add
the product of a, J, and c.
41. To the product of a and h add the quotient of x di
vided by a, and divide their sum by y diminished by c.
2
18 ELEMENTS OF ALGEBRA.
42. Write a exponent n plus the quotient of x divided
by y, minus h times the quotient of h divided by the ex
pression, a exponent c plus & exponent m, is greater than h
minus x,
43. Write x fifth power minus h sixth power ])lus y to
the ??ith power, divided by z to the n\\\ power, is less than
g tenth power.
44. Write a to the ??th power divided by h exponent m,
minus x exponent n, equals a minus h, divided by the sum
of a second power and h third power.
45. Write c fourth power divided by a second power,
minus the product of ^ and y, plus x .... written n times,
equals a exponent m.
46. X exponent m, plus the fraction, a fifth power minus
three times a second power h third power, divided by x
minus y, equals x minus y, added to the sum of 4 a and h
minus m, plus 1 divided by x to the nth power.
47. Five times the third power of a, diminished by
three times the third power of a times the third power
of &, and increased by two times the second power of h.
48. Three times x exponent 2, minus twice the product
of X exponent 3 and y, plus the third power of a.
49. Six times the third power of x multiplied by the
second power of y, minus a exponent 2 times the fourth
power of h.
50. a times the second power of n, divided by x minus
?/, increased by six a times the expression x plus y
minus z.
ALGEBKAIU ADDllluN.
19
CHAPTER ir.
ALGEBRAIC ADDITION.
14. In Art. 12 it was shown that to add a positive
miniber means to count so many units in the positive di
rection, and to add a negative number means to count so
many units in the negative direction.
In Algebraic Addition of several numbers, we count from
the phice in the series occupied by any one of the num
bers, as many units as are equal to the absolute value of
the numbers to be added and in the direction indicated
by their signs. Thus,
ExAMPLK I. Fiml the sum of 3 a and — 9a
Solution. 3 a signifies a taken 3 times in the positive series, and
— 9a aigniiies a taken 9 times in the negative series We count from
+ 3 a, 9 a units in the negative direction, and a is tiikeii ia all 6 times
in the negative serie.*?, or —6a. That is, 3a+ (— 9a) =  6a.
Similarly (} 9 a) + ( 3 a) = 4 6 a.
Example 2. Find the sum of a, 2 6, and (— 3 c).
> +
"N^
>i
+Z^
A D
c
o
3C
<
20 ELEMENTS OF ALGEBRA.
Explanation. Suppose these algebraic numbers to be accurately
measured as represented on the line of numbers A C. Start at By
then count 2 b units in the positive direction and arrive at G. Now
count 3 c units in the negative direction, and arrive at D in the posi
tive series.
Thus, (+ a) + (+ 2 6), or a + 2 6 = ^ C ;
a { 2b + ( 3 c), or a + 2h  3 c = A D.
The sum of the algebraic numbers is equal in absolute value to
A D in the positive series. That is,
(+ «) + (+ 2 6) + ( 3 c) = a + 2 6  3 c. Hence,
The sum of several algebraic numbers is expressed by con
necting them loith their proper signs.
Notes 1. The sum of several algebraic numbers is the excess of the num
bers in the positive series over those in the negative series, or the excess of the
numbers in the negative series over those in the positive series, according as
the one or the other has the greater absolute value. Thus, in Example 1 the
algebraic sums are —Qa and + %a. In Example 2 the algebraic sum is A D
in the positive series.
2. The sum of algebraic numbers is the simplest expression of their aggre
gate values.
3. Algebraic addition is not always augmentation as in arithmetic. Thus,
(+ 7) + ( 5) = 2 ; also (+ 8) + ( ll) =  4.
15. A Coefficient of a term is d^ factor showing how
many times the remainder of the term is taken. Thus,
In the term 5 abm, 5 is the coefficient of a b m, and shows that
a 6 m^s taken 5 times ; 5 a is the coefficient of 6 m ; 5 ab is the co
efficient of m. In the term 4 m (a 6  2 a), 4 is the coefficient of
m (rt 6 — 2 a) ; 4 w is the coefficient of (a 6 — 2 a).
Note. A coefficient may be numerical or literal. When no nnmerical
coefficient is expressed, 1 is always understood to be the coefficient; as, x ; xy°.
Like Terms are those having the same letters affected
with the same exponents. Thus,
ALGEBRAIC ADDITION.
21
2m^ujc^ </*, m"^ n jfi y*, and — 10 m* n x* y^ are like terms, as are
also bx^^y^z* and — 3x"y*z*; but dx*y^ and 5x*i/*2* are un
like terms. Like terms are said to be similar.
A Monomial or Simple Expression consists of one term; as,
x: lU (( l>c: — 5 a^ j^.
16. To Add Similar Monomials.
I. When all the Terms are Positive or Negative. Add
the numerical coefficients; to the sum, annex the common
symbols^ and prefix the common sign.
II. When Some of the Terms are Positive and Some are
Negative. Add separately th£ numerical coefficients of all
tlie positive terms and the numerical coefficients of all the
negative terms; to the difference of these two results^ annex
the common symbols, and prefix the sign of tlie greater sum.
EXAMF'LE 1. Find the sum of 10 a; y*,  Sxy^ 4x?/*,
and —\lxy^.
Explanation. For convenience write the terms
as shown in tlie margin. The sum of the coeffi
cients of the positive terms is 14, and the sum of
the coefficients of the negative terms is 'SI. The
difference of these is 17, and the sign of the
greater sum is negative. Hence, the required
sum is — nxy*.
Uxf,
Process.
+ Axy^
 3xy^
 Uxy^
— M xy^
— 17x2/*
Example 2. Find the sum of (x f y), 1.1 (ic + y),  2.9 (x + y),
.29 (x + J/) .  i (x + y), and 1 .26 (x\y).
Explanation, (x+y), enclosed in parentheses,
is treatt^l as a simple symbol. The coefficients
of (x + 1/) are 1, l.l, 2.9, 29, i, and 1.26. The
sum of the coefficients of the positive terms is 3.65,
and the sum of the coefficients of the negative
terms is 3.15. The difTerence of these is .5, and
the sign of the greater sum is positive. Etc.
Process
+ C^ + y)
1.1 (x + y)
+ .29(xf2/)
f 1.26 (x 4 2/)
2.9 (x + y)
 H^ + y)
+ .5(xfy)
22 ELEMENTS OF ALGEBRA.
Exercise 7.
Find the sum of:
1. (+ 2 a), (+ a), (+ 4a), (+ 3a), (+5 a), and (+ la).
2. (+ 5 a a?), (+ 2 a x), (+ 6 a x), (+ a x'), and (+ a x).
3. (+ 6 c), (+ 8 c), (+ 2 c), (+ 15 c), (+ 9 c), and (+ c).
4. (6a&c), (+4«&c), i+ahc), (2abc), and (+5a&c).
5. (fa;2), (^^2)^ (1^2)^ (i^''), and (x^).
6. (+ f ^), ( I ^), (+ I ^\ ( 2 ^), (+ I x), and (+ ^).
7. (+ 3 a3), ( 7 a3), ( 8 a«), (+ 2 a^)^ and (11 a^).
8. (+ 4a2?>2)^ (_ aH2), ( 7 a^h^), and (+ .5a2&2).
9. (+7ahcd), (+ 2ahcd), (l.labcd), and (4.1aZ>ctQ.
10. + (ft ^ 6'),  .01 {b + c), + .7{b + c), 10 (b + c),
and + i(b + c),
11. +10(,.— 2/)3, (x'2/)3, +m(xyf, 2{xy)\
and — 3 (ic — ?/)^.
17. If the monomials are not all like, combine the like
terms, and write the others, each preceded by its proper
sign (Art. 14).
Example 1. Find the sum of (+ 7 k), (^ 3 h if), ( 2 x), (5 6 y%
(+ 4 X), (8 6 /), (+ 9 x), (+ 6 1/), (+ 1 1 x), and ( h y^).
ALGEBRAIC ADDITION.
23
Ilzplanation. For convenience, write
the expressions so that like terras shall stand
in the same column, as in the maigin.
The sum of the terms containing x is
f 29 X, and the sum of those containing
b y^ is — 10 6 y"^. Hence, the result is
H 29x 10 6 2/'*.
Process.
4 7x4 3 6 1/2
 2 X  bhy^
+ 4 X  8 6 1/2
+ 9xf 6 1/2
+ llx 6y2
+ 29 X  10 6 1/2
Example 2. Find the sum of f .05 (a  6),  .01 (m + n)
4 7 (a + 6),  3 (m + n),  1 1 (a + 6), and f 10 (m h n).
Explanation, (a f &) and (m + w),
enclosed in parentheses, are treated as
simple symbols. The sum of the like
terms containing (a+/>) is — 3.95(aft).
The sum of tiie like terms containing
(m + «) is + 6.99 (m + n).
Process.
f .05(0 + 6) .01(m + «)
+ 7 (a + 6) 3(m4w)
 11 ((! + ?>)+ 10(m + yi)
— 3.95 (a + 6) f 6.99 (m + n)
Exercise 8.
Find the sum of :
1. (+ .3 X), (+ .5 y\ (+ .01 .>), (+ 3 y\ and ( 7 x).
2. (4 I a), ( J a J), (+ § a), (h ^ rr ft), and (  ..)•
3. (+ 5 (2 ./2), ( 2 a«./;), ( 2 c2.r2), (+10 a^ x), (+ 8 c2 x^),
(4a«x), (4c2.x'2), and (f 4a8a?).
4. (+ I .^), ( J a 5), (+ V •^'). (+ 1^5 ^^ ^'). (+ A^^)>
(+^^'). (HJ«^). (i^^), {i^^''). and (+§«6).
5. 7a,  :^xyl 8a, .3 (x?/), .03Cr  y), and .la.
18. A Polynomial <>i Compound Expression consists of two
or more terms.
24 ELEMENTS OF ALGEBRA.
Example. Find the sum oi 8ax— .ly + 5, .7ax + y — am — 9f
and  .3ax 1.02 y + 5p .3.
Process. Sax — .1 y +5
.7 ax + y — am — 9
.Zax l.02y ~ .3t5p
8Aax — .12 y — am — 4.3 + b p Hence, in general,
To Add Polynomials. Write the expressions so tliat like
terms shall stand in the same column. Find the smn of the
terms in each column, and connect the results with their
proper signs.
A polynomial may be regarded as the sum of its monomial terms.
Thus, the sum of the terms (f <?), (— h), and (—3 c) is « — ft — 3 c.
Hence, the sum of two or more polynomials whose terms are all
unlike is expressed b}'^ writing their terms ivith their respective signs.
Thus, the sum of a — b, c — d, and m. + n — x is a — b + c — d
+ m + n — x.
Exercise 9.
Find the sum of : •
1. 2x{y, bx+3y, — 3x — 2y, and — 4 a^ + Sy.
2. 5x + Sy+3a, —7x + 4:y — 8 a, and 2 x — 3 y.
3. 3 63^, 2 c  2 ^, 3c 71), and 4:h2c + 3x.
4. 14a + x, I3hy, 11a + 2 y, and ic  2 a — 12 &.
5. ax — 4:mn + hd, hd — a x — 3 mn, 7 m n — 3 ax
+ 3hd, and 5mn — 30 ax — 9hd.
6. a — h, 21) — c, 2 c — d, 2d — 3(f\n, and m — n + x.
7. a b c \ 3 a b m — 5 c m, 3cm+ 11 abm\9abc,
90abm — 21cm — 31 abc, and 3cm — 51 abm\ 13 abc.
8. T/i + n + p, m — n—p, m — n+ p, and m \7i—p.
ALGEBRAIC ADDITION. 25
and — a + 6 + f 4 rf.
10. 1.25 a 6 + 1.1 c + 99 h, ami 3 ah ■{■ 2.2 c \ 1.01 i.
11 ia^iah + .QV', la  ^^^ « ^ " 1^ ^'. I ^' + i^^2>
f .6 ^, and .1 a  1.01 « h
12. J ?/2  .2 wi f J, .1 wi2 + .01 m  2, m^ _. 3 j ^^^ ^^ _ ^^
and J m — 5S m 71 — 1 J w^ — 2.
13. xy — ac, Sxi/ — 9a(\ —7xy + 5ac+lMcd,
4 xy ^ 6 fl c — .09 f rf, and — x y — 2 a c + c d.
14. .5 r(3  2 r2 h  I b\ lu^b .75 a l^ + 2 b^ and
15. 3 (m  nf + .3 (a: + ?/)^ .4 (m  7/)3  .2 (x + y)\
.7(mn)83.03(a; + i/)^ and 5.1 (m 7^)3  M 1 (.aH.#. '
16. oa^V^^a^b^\x^y^xf, 4a^b^7 a^ b^'Axf
+ (jx^y, 3a^l^^Sa^b^3a^y+5x y\ and 2 a^lfi
^a^JJi ^ 3 a^ y — 'S X if.
17. aa.2§/^2 4. .,3y^. j8^ 3^.x^+ .r^?/+ 7.5^2
+ J fcs 2 ^ .^2 ^ 3 ,3 ,j _ 1 ,,2 _ 1 2,8^ j^.^a Jj. a x^V\x^ y
18. ac^ ^ \a}?' V In"^  \a^b  \abc\ \a^ c, \nn
j^lb^^laV^^ \b<?\2ahr + \\\?c, and 1.1 ^rc 1 2ac2
+ i^^c />62l l..Sc3+ 1.23 r/ 6c.
19. 3.1«84.2a?2+1.2a; + 1.7, 2.22 a,3 1.2 a^ + 3.33 x
 10.09, 2 a^ + 7 a^  2 .r + 1, 3 ./;8 + 1.22 a^» + 12.12
 1.33 a;, and 11.1 1 j? + 5.55 x^  0.2 u. + 3.77.
20. «H2c3 + «2^^,2 ^ 3 ^^2/^3 ,5 13 ,,3^2,;3 _ 1 4 ,,2^,3^8
+ 1.5 a2 63 c2, 1.5 «2 63 c2  1.9 ^3 }?• c^ ^ 1 3 ,,2 ^3 ,5 and
1.7 a3 62(^3 _ 1.2^(2^^5 4. 101 a^ftSca.
26 ELEMENTS OF ALGEBRA.
21. 2c'"+.la" + 3&c, .5c'" + 3.9ft"+2.02&c, c'" + 2.09a'»
&c, and i& + f Z^c 3.03.
22. a & — a:' +  a??/, .1 X { .01 a ic — 2.02 a 6, ^ ax
■\ ^xy — ^ah, 6ic ~ 1.01 ax + .la;, and ■J'^' 2/ + j a x
\xy,
23. 3 ^ + .ly 7.01 ^ + 6.01 ?/ + .2 a + 3 ^  1.5 c;
.8^ + 9.01a + 3.03?/ + ^4.04?/ 2.01 2;+ 2.2 a y.
24. 3 a 6 4 9  ic2 ?y, a;3 3/ + 3 a; 2/ + 5, 6 a;^/^ + 4 a;2y
— 3 icy, 10 o^y + 1 + 3 a; 2/2, and 17 — 3 x^y — 2 a;^y.
25. .5 {m  3xy,   (m  Sx)\ .75 (m  3 x^, and
— 1.25 (>yi  3 xy.
26. ^2 + Z;4 4. c3, _ 4 tt2  5 c^ 8 a2 _ 7 &4 + 10 c3, and
6 &4  6 c3.
27. 3a24a& + &2+ 2a + 3&7, 2^24^2 + 3a
56+8, 10 ab+ Sb^ + 9 b, and 5 ^2 _ 6 a & + 3 ^2
+ 7 a  7 & + 11.
28. x^ — 4:a^y + 6x^y'^  4:Xy^ + /, 4:S(^y — 12x^y^
+ 12 xy^ — Ay\ 6 x^y^—12xy^ + 6 y^, and 4:xf — 4:y\
29. a3+ «62+ ac^a^babca^c, a?b ^ b^ + bc^
— ab"^ — b'^c — ab c, and a^ c + b"^ c + c^ — a b c — b c^
— a c2.
30. 5 a^  2 a^b + 9 ab^ + 17 b^, 2 a^ + 5 a^ b
4:ab^12b^ b^4:ab^5 a'^ba^ and 2 ^2 6
2 a^Ob^ab"^.
31. a;»« — ?/"+3< 2ic'"3/a, and a;'» + 42/"— a''.
ALGEBRAIC SUBTRACTION.
27
CHAPTER III.
ALGEBRAIC SUBTRACTION.
19. In Art. 12 it was shown that to subtract a positive
imuiber means to count so many units in the negative di
rection, and to subtract a negative number means to count
so many units in the positive direction. Hence, the addi
tion of a positive number produces the same result as the
subtraction of a negative number having the same absolute
value.
Thus, 43+ (+6) = + 3 + 6 = 9. +3 (6) = + 3 + 6 = 9.
Also, the subtraction of a positive number produces the
same result as the addition of a negative number having
the same absolute value.
Thus, + 4(+6) = f46 = 2. +4+(6) = +46 = 2.
We observe that the subtraction of one number from
iinother produces the same result as counting or measuring
from the place occupied by the subtrahend to the place
occupied by the minuend. Thus,
Subtract — h from + a ; also + a from — 6.
a^
<r
b
ab
+b
a<r
IB
D
28 ELEMENTS OF ALGEBRA.
Explanation. Suppose the algebraic numbers to be accurately
measured on the line of numbers C D. We start at B in the positive
series, count b units in the positive direction, and arrive at D ; and
the distance from A (0) to Z> is equal in absolute value to A D in the
positive direction. But in counting from ^ to Z> the absolute value is
the same as the absolute value in counting from C (the subtrahend)
to B (the minuend), and we have counted in the direction opposite to
that indicated by the sign of the subtrahend. Thus,
C B = h {+b)\ (+a) = a + b. That is,
(+ a) (b) = a + b.
Subtracting + a from — b gives the same result as counting from a
in the positive series to b in the negative series, and the distance from
5 to C is equal in absolute value to B C in the negative direction.
Thus,
BC = + (a) + (~b) = ab. That is,
(— b) — ({ a) = —b — a, OT — a — b. Hence,
Algebraic Subtraction is the operation of finding the dif
ference from the subtrahend to the minuend.
To subtract — 5 a from + 2 a is the operation of finding hoio far
and in ivhat direction we must go to pass from 5 a in the negative
series to 2 a in the positive series, and is found, by counting from
— 5 a to + 2 a, to be 7 a units in the positive direction. That is,
+ 2a  ( 5a) = + 7a.
To subtract + 5 a from  2 a, we count from 5 a in the positive
series to 2 a in the negative series and pass over 7 a units in the
negative direction. That is,
 2 a  (4 5 a) =  7 a.
These differences may be found by changing the signs of the sub'
trahend and proceeding as in addition, as shown by a comparison of
results. Thus,
Minuend. Subtrahend. By Addition.
+ 2a(5a) = + 7a > j^+2rt + (t5«)=47a.
2a (+5a)=7a ' I  2 a f ( 5 a) =  7 a.
+ a  ( &) = + a + 6 f ''''' ) + a + ({ b) = a + b.
 b(+ a) =ab ) \^ 6+(a)=a6.
ALGEBRAIC SUBTRACTION. 29
Hence, in general,
To Subtract one Algebraic Number from another. Change
the sign of the subtrahend, and add the result to the minuend.
Notes: 1. Algebraic subtraction considered as an operation is not distinct
from addition ; for it is equivalent to the algebraic addition of a number with
the opposite algebraic sign. It includes not only distance but direction, and
direction depends upon the sign of the subtrahend and which number is consid
ered the minuend.
2. Algebraic subtraction is not in all cases diminution. Thus,
8 ~ ( 2)  10 ; also 2  ( 8)  10.
E.xAMPLE 1 . Subtract + 3 o* 6 c w* from f 10 a* 6 c 7/1^.
Solution. Changing the sign of the subtrahend, and proceeding
as in adtlition, we have + 10 a* 6 c m^ f (— 3 a* 6 c m*) = + 7 a^bcm^.
Example 2. Subtract + 27 (x^  i/«)» from 13 (x^  i/)».
Solution. Treating (t* — y^y as a simple symbol, changing the
sign of the subtrahend and proceeding as in addition, we have
13 (x2  7/)« + [ 27 (x^  !/«)»] =  14 (z"  y«)».
Exercise 10.
From :
1. +OaHc take —aHc, —Ual^xy^ take +19 al^xi/.
2. +:tVtake./.y; +99 mnp^rst^^ take ■}99mnphst^^.
3. —10 axy take —axy\ x'^ take —he.
From the sura of:
4. — 11 a.^, + .5 d^, and + 1.25 J" take f 5.5 0^.
5. ah^, — Sabc^, and + .3 ah(^ take the sum of
abc?^, + SMabc^ and  IMabc^.
6. lOS mnp^^, —lO.Smnp^^, and +vinp^^ take the
sum of  10 771 np^^, + 33 mnp^^, and — 108.1 m np^^
30 ELEMENTS OF ALGEBRA.
7. 5 {x + y), — 2{x\ y), and + {x\y)y take the sum of
 {x + v/), + 6 {x + 2/), and  2.5 {x + ?/).
Find the aggregate value of:
8. + 17 a x^ (5a a:3) + ( 24 a ^'S)  (+ a j:^).
9. + 19 a ^ ?/2 + (+ a ;2; 7/2) — (— 5 a x ?/).
10. + x2y + ( x^y)  (+ x^y)  ( ^2y) + (_ 1 .^^2^/)
 (+ 3 x^y)  ( 10a;2y) + {bx^y). '
11. + n« + ^)'  [ 1 (« + ^)'] + [(« + ^)']
 [+ I («■ + ^)'] + 10 (a + ?^)2.
12. I . + (+ 1 X) + ( .1 X)  (+ ,2^ ^") + ( 1 x^
+ (+ 1. ix''){2>ix'y
20. Example L Subtract 2 ah + f) a^ if  U a^  1 y^ from
15 a3  8 1/ + 23 a3 t/S.
Process.
Minuend ] 5 a^ _ 8 ?/3 + 23 a^ i/S
Subtrahend, with signs changed + 14 a* + 7 ?/^ — 5 a^ ?/^ — 3 a 6
Difference 29 a^  ^/^ + 18 a^ ^/S  3 a 6
Example 2. Subtract 3 a; ?/ + n  5 a^ 6 + 5 jo'' from 5 ic i/^  3 a^ 5
+ 3 771.
Process.
Minuend 5 xy^  3 a'^h \ Sm
Subtrahend, with signs changed 3a: 7/^+5 a 6 — n — b p^
Difference 2xy^\2a^h+'3m — n — 5p^
Hence, iu general,
To Subtract one Polynomial from another. Change the
algebraic sign of every term in the subtrahend, and add the
result to the minuend.
Note. It is not necessary that the signs of the subtrahend be actually
changed, we may conceive them to be changed.
ALGEBRAIC SUBTRACTION. 31
Exercise 11.
Subtract :
1. 5x~3yf22 from 3 x \ y — z.
2. X — y •{■ z from — x — 2 ij — 3 2.
3. a — 6 4 20 6' from — a — h •{• 10 c.
4. J^ — iy — i^ from \x \ y + z.
5. lx\y\^zhomlx+t^y^^z,
6. J^iyi from JxJy+ J.
7. a3 _ 4 a2 J ^ 2 i c + 5 from 3 a^  «U  5 c  5.
8. a^y — Z ah X \ 2 X y^ — \ from S x^y — abx^ 2 xy^.
9. ahcxy \2ahy — 4ihx from 2abcxy — ab y \ bx — 'S.
10. acy — bxy + aJc — 1 from bxy — 4:acy — abc fa.
11. .4a:*.3a:3^.2ic27.1a; + 9.9 from a;^  2.10 a:^
+ .2ar»:.07a.' + .9
12. 1.2 /s  1 4 X + 1.1 a;4 + 1.7 a^ + a from \x
.l.r* + .2.'/^.3A'8 4a.
13. \ vin^ ^In^ yi^ + I w2 from  /7i3  J r/i 7i2  ;,2
14. .125 m3  M^ ,n vi2 _ .33^ ^ ^2 _ (5o ^3 _^ ^. Vom
g Wl^ — I Wl n^ — I ?//,2 7J, I J 7^3
15. J ,;i2 _ I y _ I ^ 4. J ^ .^Qj^ ^ W2 ^ y 4 f ?l  i .^.
16. a^hc + xf Ic from 3^ a2ftc + 2Jaj?/" 4f c.
17. lOc — ah ^ b(l\ 6rt15c4 3(Z from 25«5
— 5c48r? — 20rt.
18. af" — 3 a:" if — if from 4 x*" + .c* y" _ af .
32 ELEMENTS OF ALGEBRA.
19. — .9 a"* x^  .3 a 63 a? + .6 + .03 IT c x^ from .9 a'^x^
+ 1.3 2ah^x\ AlTcx^
20. From the sum of lo?\h^ +\ c\ J cv^  3^ c\
and 2 63  I a2 _ i c^ take ^i a?  ^\ 634 c\
21. From the sum of 3 ic3 _ ^2 ^ _ ^^^ 7 ^3 _ il ^2;3
f llj y^ oc, and 11 ;;c2;3 _ gi ^3 _ 2 1/2 a; take a;2;3 _ 25 ^/^j;
4 1 ^3.
22. From of + y"" take the sum of 11 ^^ + y^ — z,
— 6 ^" — 5 /* — 3 ;2, and  5 ^" + 3 2/"* + 4 z.
23. Add the sum of S^y .3 ^/^ and 53y + 2.7 2/3
to the difference obtained by subtracting 3 + l^y^ — .5y
from 1—2/3.
Queries. Why change the signs of the subtrahend in subtracting ?
Wliy add the subtrahend, with signs changed, to the minuend ?
Does the use of the signs + and — in Algebra differ from their use in
Arithmetic ? How ?
Miscellaneous Exercise 12,
1. From m^ — n — 1 take the sum of 2 n — 3 + 2 ?^3
and 3 ??i3 — 4: + 5 'n? — n.
2. From the sum of 1  8.8 y + .9 a^ and 1.1 x^ + S x^
— .2 2/ — 1 subtract 2 x^ — x^ + 5 y.
3. Take x^ + x — 1 from 2 x^, and add the result to
2a^x^x + l.
4. Take a^ — h^ from ah — h^, and add the remainder
to the sum of a6  ^2  3 6^ and ^2 + 2 62
5. To the sum of m + 7i — 3 /? + 5 and 2 m + 3 ?i + 5^
— 3 add the sum of m — in — 7 p and 5 ^ — 6 ??i — 2.
ALGKBRAIC SUBTRACTION. 33
6. Take 3 a^"*  2 x2" f^  y^^ from 3 f/—' + 2 x^'^y^
7. Take 2x^y^*6z^\2y^V z1 from a;iy*+2;4yi
8. Take .8 h^ y^  A a? x^ + .3 rf from .4 a? x^ + .3 c
 1.2?)tyi
9. Take f xtf a;*y*H 33j?/§ from f j:;^ + ^*?/* + Jyi
10. From the sum of .7 c y^ — .4 a x f .5 h, .04 6 —  cy»
+ i w,a.r+ Jcy§, and yi a x .23 6 .8 m + .3
take the sum of .55 a ic + J /?i + ^^ and .33 in — 1.1 cy^
+ .67 ?«.
11. Find the sum of «"•  7 6* h cp and  6" +  a"*,
and subtract the result from cp — 4:n.
12. From a*  2 c"  af* take the sum of J «"•  ^ i"
 X' and i a"* + J 6" — y" — af.
13. From  a*6*c'^« take the sum of Jft'"+ §6*
 I C, ^5 a"*  \l c^, and V ^^ + ^'■
14. From 3 (a^  ^y  (o^a + ff take § (^ + 2^)*
 c^ + 3J («^  ^)*.
15. From unity take 3 a^ — 3 a + 1, and add 5 a^ — 3 a
to the result.
16. Add 3 ^  7 a:" + 1 and 3 a;8 f ^  3, and
diminish the result by x^"* — 2.
17. From zero subtract I a^ — ^ x \ 2.
18. From .3 m'  1 + J w take 5 n^  2.7 m^  J n,
tlien take the difference from zero, and add this last result
to  5 n^ 4 3.3^ m^ f n.
34 ELEMENTS OF ALGEBRA.
19. What expression must be subtracted from 10 y'^ ] y
 1 to leave 3 y^  17 y + S ^.
20. What expression must be subtracted from a — o x
+ y to leave 2 a — o x \ yl
21. From what expression must a^—bah — lhc be
subtracted to give a remainder b a? \ 3 ah \ 1 h c1
22 From what expression must a^ h^ — b^ c^ + 6 a"" c**
be subtracted to leave a remainder b^ c^ — 6 aJ^ c" ?
2*3. To what expression must  ft^ + 2\a — 1^ a^ — 3
be added so as to make 2\ a^ — 2\ a \ 3\ a^ \ ^ ^
24. To what expression must b x y — lib c — 1 mn be
added to produce zero ?
25. What expression must be added to 3 ^" — 3 ^""^ + 2
to produce ^" + x^"~^ — 6 ?
26. What expression must be added to m a?"* — ^" + 2
to produce m a?'" — 2 ?
27. From the sum of .6 {x + y)^ + .3 a" + ^'",  a** x"^
— c^ ~ I (« + ?/)2, and I (^ + 2/)^ ~ I ^" ^''"j take the sum
of .3 (x + i/)^  I a" ;:c^, I «" ic'"  6.5 (ic + y)^ + c^ and
ro (^ + .?/)^ + 3.3 a" ^"^ — 3.
Algebraic Subtraction may be defined as the operation of
finding a number which added to a given number, will
produce a given sum. The sum is now called the min
uendy the given number is the subtrahend^ and the required
number is the difference.
ALGEBRAIC MULTIPLICATION. 35
CHAPTEK IV.
ALGEBRAIC MULTIPLICATION.
21. Evidently dm x 6n = 5 x (^ x m x n = SOmn.
Hence, in Algebra, the product is tlie same in whatever
order the factors are written.
a X a X a X a or aaaa is written a*, and shows that
a is taken four times as a factor. aXaXaXaxaov
aaaaa is written a^, and shows that a is taken five times
as a factor, a X a X a X ton factoi*s, or a aa to ?i
factors is written a", and shows that a is taken n times as a
factor. Hence,
An Integral Exponent shows how many times a number
or term is taken as a factor.
a^ is read a second power, or a exponent two, or a square.
a' is read a third power, or a exponent three, or a cube.
Hence,
A Power is the product of two or more equal factors.
The degree of the power is indicated by an exponent.
a? = a a a,
and a^ = a aaaa a .
Hence, a^ X a^ = a a a a a a X aaa
= a»
a" = a a a a .... to n factors,
and a*" = a a a a .... to m factors.
Multiplying the second expression by the first, we have,
rt"* X a* = aaa to m factois X a a a .... to n factors
= a a a .... to (m + ?i) factors
= a*"*"". In which m and n are a?iy numbers
36 ELEMENTS OF ALGEBRA.
whatever. Similarly for the product of more than two
powers of a factor. Hence,
The i^owers of a number are multiplied hy adding the
exponents.
If the multiplicand and multiplier consist of powers of
different factors, we use a similar process. Thus,
3m^ X 2 111^11? X 5 m'^n^ = o X 2 X 5m 771 m m m mmmnim
X n n n n n
a"fe"* X a^lf =aa a .... to n factors x aa a .... to p factors
Xbbb to m factors Xbbb .... to r factors
= aaa .... to {n\p) factors x bbb .... to (m + r)
factors
=: a""*"^ 5'" '^'■. Hence, in general.
To Find the Product of Two or more Monomials. To the
product of the numerical coefficients annex the factors, each
taken with an exponent equal to the sum of the exponents
of that factor.
Notes: 1. When no exponent is written, the exponent is 1. Thus, a is
the same as ai, & as fti.
The exponent is used to save repetition.
2. We read a^, a square, and a*, a cube, because if a represents the number
of units of length in the side of a sqiiare, and the edge of a cube, then ffl2 and
a3 will represent the number of units in the surface and volume of the square
and cube, respectively.
Illustrations.
11 mi9 X 10 m}^ =11 X 10 mi^ + lo = no ni^^.
3a'^bcmX2ah'^cmX 5abc^m^ = '3 X 2 X 5a2+i+i6i+2+ici+i+«mi+i+2
= 30 «4 ^i c* m\
3a^fe*c3 X 4rt X 6'cJ = 3 X 4 a' + ift^ + J c3 + ^ = I2a^bc*^\
S'x^i/" X 2^x3 X x^y"" = 25 + ^x23+57/" + " = 2 x*i/2".
ALGEBRAIC MULTIPLICATION. 37
Exercise 13.
Find the product of :
1. 0? and 7 a?^ ; 3 a a; and ^ c^t^ \ a? ha? and 2 a^ W a?.
2. ^xyz^^ and Is^t/^mn; ^abcdm^7i^ and ^a^lf^c^d^mn.
3. ^/2^.3yaud a6366y" + '; 3a3a:«3/7andf ai«^a:8 3^*2.
4. 3 a a:^y^ and 10 a^^xy^^ ; J x"* if and  rr* y^.
6. 3aftca;^V and f rt^j^c^^^ \a'hh^xy and  a Z>iOc V^^"*.
6. f a'fcaf / and .2 a^ h^ 2^ f , a? f and a^^i/S.
7. a'izand tt6'"; a^^V^^and 5.7a:i3/i.
8. ahx]/^ and a?}?x^y\ a"'+'i>'* + '* and a'"''/;"'".
9. .55a;' + '//'' + ' and .5 a;''+«/''; .3 a2"';j;8 n ^j^^
a* 2:"; 5aH".r^ and ^oa\hx'^\
10. 2«^2:, a?/, «2//, a^j^i/, and a ^.
11. a*, fc", 3c', a', 5', c*", and c?'.
12. 2c'"</, «c", r/22:", a"* a;*", and ci
13. ai VI x^, a^ ni x^, a m x y, and 2 a^ 71^ x^ if.
14. a^z^, «^?/^, ala;"i, a^y~^, 5fa"la;^, and 5^2:^yV
15. fa^ma:", jTnta:*, .Sarr"*. 5.1 w', and a^'x',
16. 3y*, a^y'^z^, (^ b"", a^b', f «^.7/", and ^a^J"'^.
17. (a 4 ft), 5 (« 4 6)2, 3 (« + 6)^ ^ (^ + &)^ and (a + bf.
18. (a + 6) (c + (if, {a + bf, 3 (c + df, and (a + 6)7 (c + rf)2.
19. 3 (a + 6) (a:  y), J (a + 6)«, and  (a:  yf.
38 ELEMENTS OF ALGEBRA.
22. Algebraic Multiplication is the operation of adding
as many numbers, each equal to the multiplicand, as there
are units in a positive multiplier ; it is also the operation
of subtracting as many numbers, each equal to the multi
plicand, as there are units in a negative multiplier. Hence,
The multiplier shows that the multiplicand is taken so
many times to he oAded, or so many times to he suhtracted.
Thus,
(+ 6) X + 4 = (+ 6) + (+ 6) + (+ 6) + (f 6) = + (+ 24) = f 24
(_ 6) X + 4 = ( 6) + ( 6) + ( 6) + ( 6) = + ( 24) =  24
(+ 6) X  4 =  (+ 6)  (4 6)  (+ 6)  (+ 6) =  (+ 24) =  24
(6) X4= (6)(6)(6)(6) = (24)=:f 24.
■ The sign of the multiphcand (6) shows that the product (24) ii^ in
the positive and negative series of numbers, respectively; and the
sign of the multipl ier (4) shows that the first two products are to be
added and the last two are to be subtracted. Hence,
The sign of the multiplicand shows what series of numbers
the product is in, and the sign of the multiplier shows what
is to he done ivith the product.
Law of Signs. The product of two factors is positive when
the factors have like signs, and negative when they have un
like signs.
Since
2x2 = +4; 2x2x3=:h4X^3==12;
2x2x3x412x4 = + 48;
_2x2x3x4x5 = + 48x5 = 240;
etc. Hence,
The product of an even number of negative factors is posi
tive; of an odd number, negative.
ALGEBRAIC MULTIPLICATION.
The change of signs may be illustrated as follows
39
> +
axSr^
K.
ox+y
»^y»w.s
ax3
V
QX^3
VCATT
ClX2
<
ax+i
Q X/
^
ax+/
^
A *
k
QX*Jl
ax+/
.
,
ax A
fly3
CtXi'
^
7 ... 
ffX+J^:
•CTK^S^
k ■
axs

< z^
Let the measuring unit be represented by a.
From A (o), the startingpoint on the scale, measure toward the
right and left. The products of + a and — a by the factors from f 5
to — 5 are :
aXl5, aX+4, ax+3, rtX42, ax + 1;
a X  1, a X  2, a X  3, rt X  4, a X  5 ;
aX+5, rtX44, ax+3, ax+2, aX + 1;
aX1, aX2, aX3, ax 4, aX5;
respectively. The directions taken by the products are shown in the
figure.
ninstrationB.
x«y» X x*z X ^j/z^ X ^xz^ X  4 yz^ = f f X ^ X 4x«j/*2»o
X^y* X  fxyZ X  yz' X  X~^'* = ~ ajm + nSnyn + n+ljlr
Exercise 14.
Find the product of :
1. 5 a, — 3 /), 7 c, — 2 a*, — 11 a^, and a; a^x, —ay^,
a a^, and — xy.
2. ahx, —ay^, —a X, and a^a?\ —al^, —hc^, —cd?,
— a, — a^, — a^, and — 5 a*
40 ELEMENTS OF ALGEBRA.
3. — a, he, — 1, ^, 1^ a^, ^x y, and 75 a; ax, ex,
— m X, — 2/**, and .3 ?/i.
4. ^abc, —d, ax, —1, and ^axyz; a'^af, af^y*", a'^V,
and a b.
5. —a^x, 3x, ah^, ay, az, and axyvw; axy, —^a^V,
and  SJa^&'a:"*?/".
6.  a'^hc, 2 h'^cd^  .5 a^ccl^,  f^ a^H^^c^^d'^^,
and a h^ d*.
7. 1, a 3, a^7^ a 0^5, aio^3^ aH^a^, and Mr/2
8. aaP, — a\ — 1, .3 a x, and — a^^/^; ^^^.^ _ ^^ ^,
and — a^ a^i.
9. — my, mx, — mn, — xy, and it* 3/3; 3 aa Jt and
— .7 «i Z>i
10. a", a^", a^", a^**, and a^". Express the result in
two ways.
11. 2^ij'x^, mx'^y'^', 3" ^"2:!, and 2^po(^f.
12. 32", 23« X 3'^^^ 32«,  23" X 3*«, S^" x 2«, and
— 26" X 3« + i.
23. Example. Multiply a + & by m ; also a — fe by m.
The symbol (a + h) m means that m is to be taken (a + V) times.
Hence,
Process.
(a + h)m — m \ m ■\ m + taken a + h times
= {m+m+m+ — taken a times) 4(7w + m + m+ — takenft
times)
~ am \l)m. (1)
ALGEBRAIC MULTIPLICATION. 4l
Also,
(a — h)m = m\miinh .... taken a—b times
= (m + 7» + mf taken a times) — (m + m+m+.... taken K
times)
=:(rnX a)(mX b)
— am — bm. (2)
Similarly, (a + ft — c) m = a m f 6 m — c m.
These results are obtained by multiplying each term of the multi
plicand separately by the mnltiplier. Hence, in general,
To Multiply a Polynomial by a Monomial. M%dtiply each
term of (lie muUiplicaiul by Ute multiplier ^ and add the
resiUts.
Exercise 15.
Multiply :
1. hc^acab hy abc; S aH^  ^ hh^  ^ c^ hy f^a^b^^.
2. 5a^b'2c^ by a^b^c^^; .6 3^ .5 2^y^ .32^y^
 .2a^ by .2x^f.
3. j W.2 — ^mii { ^n^ by ^mn; x — y — ^x^y^ by xy.
4. fa^j&2_^^^a62 by«Z^2. a'  a^lr^ab hy ah^.
5. 6a2a3 .5a^b^x^^ \ .2h^2^ by ^ ab 3^; pxTqx*
— r by p2^ r.
6. 3a'"»2 6"H4a'"6" by a&2; .4a— " 5'''Ja*'6'
+ ft3^ by I a^^'ft'.
7. a*3a'"i—4&'* by a'J'^"; 2^2:^  2iyi + 2^a;iyi
by 2^x^yi
8. a?a2jfan + 5iby aU^; a;*2rty^4a:^i^t.6y*
by xhj~^.
42 ELEMENTS OF ALGEBRA.
Find the product of :
9. x^y^4: x^f+ 4/, xh/, and 2i/; m''^  2m^^7V''
+ n^\ m~^, n\ and in" n\
10. ^^264 + 1^632: + 1 62 2:2^ laV^, %b^x, and ^aH^a^.
11. a;3 — 7/F a;3, 2/3^ and —x'^i/^; a—M, a?, 63, a^b^,
— a2 jf and — a b^.
12. x^—i/, x^, x^y^, —x^yi,  ?/f , J 2:t, —  ?y^, and
.21 .2_1_
rr^ 7/4.
13. J  .2 6f :2:2 + ;3 7; ^i _ ^1^ J ^1^ _ il ^2^ and J b^ xi
14. 1^^'" y «6"' + 6, 'Sa"\ rjb'^, and
15. «"+« + f/"6'« + (fc"»6'* + 6'"+", a.'", ft"*, rr", 6"", and
24. Example 1. Multiply m{ti by a;+y; also m^n hyx — y.
(m + w) (v+y) means that a: + ?/ is to be taken m + n times. Ifence,
Process
(m + n) X 0 Ty) = (^' + y) + (^' + 2/) + C^' + y) 4 • • . • taken w + n times
= [(*' + ?/) + (•^■ + ?/) + (x + y) + ... taken m times]
+ [ (*■ + ?/) + C^' + y) + (^ + y) + taken n times]
= (x { y) m + (.r + ?/) n
= (1) Art. 23, ??* a 4 m y + n x \ n y. (1)
Also,
(m + n) (x  y) = (xy) {■ (r~y) + (xy)\ .... taken w + n times
■ [(^ — ?/) + (t ~ ?/) 4 (?' — ?/) 4 — taken m times]
4 [(^ " y) 4 (j" — ?/) 4 (r — 1/) f — taken n times]
= (x ~ y) m{ (x  y) n
— (2) Art. 23, nix — my \ 71 x — n y. (2)
Similarly, (7n \ n + p) (x\ y — z) = m x + 7n y — mz + nx + n y
— 71 z \ p X j p y — J) z
ALGEBRAIC MULTIPLICATION. 43
These results are obtained by multiplying each term ot the multi
plicand separately by each term of the multiplier, and connecting the
products with their proper signs.
Example 2. Multiply j*i*+2x^x5 by x*h3x*\ 5.
Process.
x« r«j2x«a:5
X* + 3 x« + 5
x"  x« H 2 x«  x«  5 X*
+ 3z» +6x*3ir*~3x8 i5x«
+ 5 x«  T) j;5 + 10 x**  5 x  25
xW H 2 x» I 7 x«  « X*  3 x8  16 x» + 10 x2  6 X  25
Explanation. Multiplying each term of the multiplicand by
each term of the multiplier and connecting these results with their
proper signs, we have x**' — r* f 2 x* — x* — 5 x* + 3 x* — 3 x* 4 6 x*
 3 X*  15 x« + 5 x«  5 x« + 10 x2  5 X  25. Umling like terms,
for a simplified product, we have x^<> f 2 x* — 3 x^ f 7 x* — 8 x* — 15 x*
+ 10 x«  5 X  25.
The process used in practice is shown above. The first line under
the multiplier contains the product of the multiplicand and x*. The
second contains the product of the multiplicand and 3 x*. Etc. To
facilitate adding, write the several products so that like terms shall
stand in the same column.
Hote. It i^ convenient to arrange the terms of the multiplicand and multi
plier according to powers of some common letter, ascending or descending.
Example 3 Multiply f a x + f x^ + ^ a* by f fl^ + § x^  f a x.
Solution. Arrange the expressions according to the descending
powers of r. Taking the multiplican<l  x* times, we have x* + a x'
f ^ rt^x^. Taknig it —  n r times, and writing the proiluct so that
like terms nhnll stand in the same column, we have — ax* — a*x*
 1^a*T, Again, Uiking it  a^ times, and writing the pnnluct as be
fore, we have ^a^ x^ h ij^a* r + \ a*. Adding the partial products,
we have x* I ^ a*, or arranging alphal>etically, { a* f x*
44 ELEMENTS OF ALGEBRA.
Process,
f x2  f a a: + f a^
— ax^ — a'^ x"^ ~ ^ a^ X
+ ia^x^ + la^x + ia*
Example 4. Multiply  Sx^' + ^y^  .Sx^^^^ y'' + ^ + 3.3j:"'^ + ^
by — .2 j:™«/«2 + 4 a;'"' i/«^
Process.
3.3 a;*"*/" + 2 — .3 x"" + ^ y'' +^ —3 x"' ^'^y''
4 x™  1 ^'»  ^ — .2 a;"» ,y"  ^
13.2r*"»i/«+i 1.2 a:'"**/'"  12.00 x^^ + 'j/ani
 .66a;2'«?/2»,_^ 06a;2"» + i j,2«i^ g^zm+a^^ng
13.2 j2m1^2n + l_l 86^:2 '"y^» 11.94x2"' + ! ^2«I_^ (5 a,2m + 2^2«2
Explanation. Arrange according to the ascending powers of x, as
shown. The product of the multiplicand by 4x'"' ^" ~ ^ gives the
first partial product, as shown on the first line under the multiplier.
The product of the multiplicand by — .2x'"^"2 gives the second
partial product. Taking the sum of the partial products, we have
the product required. Hence, in general,
To find the Product of two Polynomials. Multiply the
multiplicand hy each term of the multijplier, and add the
partial products.
Exercise 16.
Arrange the terms according to the powers of some
common letter, and multiply :
1. r/2 J^h'^ah by « 6 + ^2 + ^2 . a'^2ax ^ 4.x^
by «2 + 4 .^2 + 2 a a^.
2. x^ \ y^ — x^ y"^ by x'^\y'^\ x + y { x — y by x^y
x + y.
ALGEBRAIC MULTIPLICATION. 45
3. y— 3 + /y2 by 2/9+2/2; a^tj — aziij^ — a^ by y^ a.
^ J a;2  I X  f by J a;2 ^ I ^ __ 1 . 1 6 «2 4. 1 2 a 6
+ 9&2 by Aa.:n,
5. x^— i/\ X — fj by x^^ y^\ x — y \ ^s^ — ax — ^a^
by \ x^ — ^ax \ ^a^,
+ \d^ by 2 a;^ + « 2: — ; ^3.
7. n7^Dx'^x^^2^x + 2hyx^2x2,
8. 3a22a3_2a + l + a* by 3ft2+ 2^3+ 2a + l + ^*.
9. 1.5 2:8 + 1.5 2^2 + .5 a:* + .5 2: + 2:^ + 1 by a^  .5 a;
+ 1 + a:*  .5 2:8
10. 1 + 9 a + 5 a3 + 3 a* + 7 a2 4 a^ by 4 a2  3 a8
4 a* 4 4  4 a.
11. 4 2:2^24. 82:^3+16/ + 2a:3y _^^ by 2:  2 y.
12. 2^12 a:3^6 4. 2^6^ 2:87/24. y8 by y« 2:8. 242^2
— 3^ y — X i^ \ x^ \ y^ by x \ y.
13. rt2 _^ ^2 _^ ^ _ ^ 5 _ rt c _ ^ c by rt + & + c.
14. ^2 _^ ^,2 _^ ,.2 4. J c + a r  a 6 by a + fe  c.
15. a6 + crf4ac46c^ by abhcd^ac — bd.
16. i2 4. y2 _ 3 3^ _ y2 by 2 a: + 2 3^  2 (2:  y).
17. 3 (m + n) — .1 X (a + h) by a  b ^ .1 (m ^ n).
18. a2f^\bx''\r by a2:* + 62;"+r; 2;^ + yi by 2:^ — y^
19. a + 6" by a'" + b"; oT 4 6* by a"  6*; 2:2 _,_ j by
a:i + bl
46 ELEMENTS OF ALGEBRA.
20. Sx"^^  2/' by 2x S'f; ax''' + &^"+ ahx
by a x^ — bx^ — 1.
21. 'Sa^^'x+'Sa^y + a''' by a'" a" + 2:; x^yi by
2^2 — y.
22. .2ai.3&t hy.2ai+.3bh^xi + xUjihy^ hy x^yk
23. a;^ ?/~t + y~^ + x^y"^ + a;^ by x^ — y~i.
Find the product of:
24. 1 + 2;, 1 + 2;^ and 1 + x^ — x — a^.
25. a; — 2 a, X — a, x + a, and a: + 2 a.
26. 3 2; + 2, 2 a:  3, 5 :z:  4, and 4 a;  5.
27. ^2 — :r + 1, a;'^ + :r 4 1, and x'^ — x'^ { 1.
28. rr^ — a x{ a?, x^ + a x ■} a?', and x^ — a'^ a;^ + a*.
29. « + 6, a  6, 3 a + &, and a^ 2o?h  a}?' \ b^
30. rr + &^ «"*&", ft''"+a"*6"+?)'", and a2'«_,^«^H + ^a«^
25. A Binomial is a compound expression of two terms ;
SiS, a — b; ab + 2b\
In each of the following products, observe that :
2a; + 3 2 a: + 3
2x + 5 2a;  5
4a;2+ 6 a; ~ 4x2+ ^^^
10 a; +15 10 a; 15
4x2 +16 a; +15 4^2 4 a; 15
2a;3 2 a; 3
2x + 5 2a; — 5
4a;2 6 a; 4^2 6 a;
10 a; 15 lOx+15
4x2+ 4 a; 15 4x216x+15
ALGEBRAIC MULTIPLICATION. 47
I. Thejirst term is the common algebmic term of the binomials
multiplied by itself, or the square of the common algebraic term.
II. The second term is the al«;ebraic sum of the other two terms of
the binomial expretssious multiplied by the common algebraic term.
III. The last term is the algebraic product of the terms which are
not common to the binomial expressions. Hence,
To find the Product of two Binomials, having one Common
Algebraic Term yidd toy ether the nfpiare of the common
tertn, the abjebraic aum of the other two tervis multiplied by
the cmiimaii terniy aiul the algebraic proditct of the terms
which are iwt common.
In general, {x \ a) (x ±b) — 2^ + {a ±b) x ± ab (1)
(xa){x±b) = x^\{a±b)x^^ab (2)
In which a, b, and x represent any numbers.
Hotel: 1. It is of the utmost importance that tlie student sliould learn to
write tlie products of binomial expressions rapi<lly, by inspection.
2. To square a monomial, multiply the numerical coefficient by itself ^ and
multiply the ejptment of eacli letter by two. The proof is evident. Thus, the
square of 2 a* 6» = 2 X2 a* ^ * A* >< «  4 ofta.
Also, (36"ar«)a = 3 X 36»x2a*« ^a  962««a;2m.
Examples. Write the product of the following by inspection :
(2 a; + 7 y) (2 I  5 i^); (a  9 6) (a  8 6) ; (a  6) (a + 1 3) .
Solution. Squaring the common term, we have 4x^. Taking
the algebraic sum of the other two tenns, + 7 y and — 5 y, we have
+ 2y. Multiplying this sum by 2 a;, we have + 4a:y. Taking the
algebraic product of the terms not common, + 7 y and — 5 y, we have
— 35 y*. We thus obtain 4x^ i 4 xy— 3b y^ for the product.
Similarly, (a96) (aSb) = a«+ (9686) X rt + (96) X (86)
= a317a672 6^.
Also, (a6)(a+13) = a2+(6 + 13) X af (6) X (413)
= a2 + 7a78.
48 ELEMENTS OF ALGEBRA.
Exercise 17.
Write, by inspection, the products of the following :
1. (a 3) (a + 5); (6+6)(&5); {x + 4) {x + S)
{x  4) (0^ + 1) ; (^  7) (x + 2).
2. (x  8) (^  6) ; (a + 9) (a  5); {a 8) (a + 4)
(2x4:) (2 a^  5) ; (3 ^' + 7) (3 ^  5).
3. (0^337/2) (^3_ 4 2^2). {x7y)(x + Sy);{a"^l){a+2)
(3a:55)(3a:^4).
4. (2 a2 2/3 + 4) (2 ^2 f _ 8) ; (3 a a;  4) (3 a 2: + 7)
(a:3 + 3 a) (a;3  4a;) ; (ai^  3 a2) (a;^ + 2 a^).
5. (2a; + a)(2a;2a); (2:c"+ 5a)(22;"3«); {Sx2y)
(S X + y) ; { 6 m + 2 2^) {4.m + 2 x^).
6. (:ra)(2:5a); {a5b)(a + Sb); {a^2x){a^6x);
(5xio+3a2)(5:z;io_4a2).
7. (32/25a:^)(2 2/25:?;5); (3 a^ + 2ab)(3a^4:a¥);
(a" + 3) (a»  b).
8. (4 a + 6) (4 a  c) ; (2 &  5 a) (2 c  5 a) ; (a y 4 i^;)
(a.^ + i^); (afl)(al+).
9. (2 2:^ + 1) (2 xi 4 12) ; (2 a^  3 ax) (2 a^ + b) ;
(x .Sx^y''){y .Sx^y"*).
26. (:r47/)(2; y) = x'^ + {y  y) X x ] (i y) X (y)
= 2^2 — ?/2_ Xn which a; and y represent any two numbers.
Hence, in general,
To find the Product of the Sum and Difference of two
Numbers. Take the difference of their squares.
ALGEBRAIC MULTIPLICATION. 49
Examples. Find the product of (2 a"* f 3 b") (2 a"»  3 &*) ;
(8;)* 4 ll2*)(8;>* Hz*).
Solution. (2a« + 36") X (2a"  3 ft*) is the square of
2 a*", or 4 a* •", minus the square of 3 6"", or 9 6**. Therefore,
(2 a"» + 3 6*) (2 a"  3 6*) = 4 o* »»  9 6 «*.
Similarly, (8/)* 4112*) (8p*  11 z*) = 64/)«  121 z.
Exercise 18.
Write by inspection the product of the following :
1. (2x^'Sy)(2x^Sy); (x { 2ij){x2y); (5 + 3 a;)
(53a:); (5a:+ 11) (5 2; 11).
2. {2xh l)(2a;l); (2x+ 5){2xb); (5xy + 3)
(5 a; y — 3) ; (c f a) (c — a).
3. (c2 + a2) (c2  a2) ; (m n H 1) (wi n  1) ; (« y^ 4. j)
{af }))■ (a2r2+ l)(a2a:2_ 1)
4 (a:* + 7/) (ar*  /); (1  pq) (1 f pq) {m  n)
(m^n)\ (a"* fa") (a* a").
5. {bxr^^^y^{oxy^V4.f); {h^+^f)(^2??>f))
{2!^Zx){j^+ 3a;).
G. (2aa:4 fey)(2aa;6y); (m"' + 7i») (m"' 7^»);
(10 a"  13 6—) (10 a— + 13 6—).
7. (mi + rA) (mi  ni) ; (4 ai 20a;io) (4ai + 20 a:iO) ;
(ai6f)(«i + 6i).
8. (11 a:i + 30 ?/*) (11 a:*  30 y*) ; (15 a2 6^  16 a* 6^)
9. (i«62+56Ja:i)(Ja62i6ia;i); (a+6)(a6)
50 ELEMENTS OF ALGEBRA.
10. (ah+l)(ab~l)(aH^\l); (2a'"+ 4a")(2 a"'4a")
(4^2"^+ 16^2").
11. (5 a^ + 6&2) (5 a^  6h^) (25d^ + 366*) ; {a^ + aH^)
12. (rcl + xUj) (x^  x'y} (^x^ + x'' f} ;
(f cr + i If) (f c  6«) (f f c'^ + }f 62").
Queries. In finding the product of monomials, why add expo
nents of like factors ? What is the product of a^ and a^ ? Prove it.
Why is the product of an even number of negative factors positive ?
How prove (1) and (2) Art. 25 %
Miscellaneous Exercise 19.
Multiply :
1. 2 ^2"  a" + 3 by 2 a2 + a"  3 ; 5 + 2 a;2« + 3^
by4^«3^2a
2. ft^ + 2 a^"  3 by 5  J a" + 2 ^2* ; J a;i  5 + 8 a;t
by \x^ + lx~'^.
3. 3 ^1 _ a  a^ by f a^ + a"!  6 ai ; 2 a*""^ &"'
+ a'^ 6^ by 3 a'" ?^'^  ««^ Ir^K
4. ^x''if—'ix''y^ by 4 ^«?/ + 5^2a^26. ^t" + «t"
by ai" + «~^".
5. .3 ft*  .02 ft36 + 1.3 «2^2 + .5 a2,3_ 1 2 h^ by .3^2
 .5 a &  .6 2>2.
6. 1  2 ^^  2 ^i by 1 — .T6 ; al  8 «t + 4 a^  a^
by ia~^ \ a \ I a\
7. 2x^x^3x^ by 2 x'^  3 x~^  x^ ; a"  1
+ ft " by a^ + ft~i
ALGEBRAIC MULTll'LICATlON. 51
8. ^■""■'■^  x" + '  X+ x"' by 0;" + '^  jJ"  ^; + 1 ;
a,'»+3a;"*2a,"^ by 2 .x' + ' + it: + »  3 x*.
lU. 3^ 2a;'" + '  5a;'" + =»+af* + ^ by 3a;"3 + 2 3;"*
11. a:" + ^ 3x+*+ x*+'' 2ic + * by 2a;'"+ Sic^' —
12. 5a;"V*'2a;^/^'u;**'/+' by 3 a;*+ */"'
13. //«.' + ' — 3m'*?i+m'^7i2—?/i'^t^ by m*"^— 3m?i
14. 2.«;"+V"'43a=*'"'/'"^a;"+Y'"' + 4x«+V'"'
by 2./;■+y'**'3^•^Y'' + .tV"'' + 4x"y^
15. x+*y* + x'^*if'  2a; + V~"  4a: + 'y*
+ 4a;»— y^"
16. (y+ a;)(y .»:'") ; (i^^r f •^"VXi^rHf^c/^).
17. (x'' + y~)ra:'*y); (a:i  5) (xi ^ 4), (7 x ^ 3y^)
(7x+3rO.
18. (4 X'  5 x^) (4 a; + 3 x^) ; (f A b'^  ^^ a^I)^)
(cUt + ^a^65); (a 4 7 + 3a'')(a~ + 73a'').
52 ELEMENTS OF ALGEBRA.
CHAPTEE V.
INVOLUTION.
27. Involution is the operation of raising an expression
to any required power.
hivokition may always be eflected by taking the expres
sion, as a factor, a number of times equal to the exponent
of the required power.
It is evident from the law of signs that even powers of
any number are positive ; and Ihat odd powers of a number
have the same sign as the number itself. Thus,
(— m* uY = (— m* u) X (— m^ n)
( m* 7i8)8 = ( m*?i8) X ( w* n^) X ( m* n^)
= m< +4 + 4^8 + 818 =:ml2u».
( 3 m8 ny = ( 3 m^ n) X ( 3 m^ v) X ( 3 w^n) X ( 3 m^ w)
^ + 31 + 1 + 1 + 1^8 + 3 + 3+3^1 + 1 + 1+1 = 4. 81 mi'^n*.
(a** 6<^)~ = a*^ b" X oT^ ¥ X a"^ b*" X ton factors
= (a*" X a"* X a"* X .... to n factors) X (6^^ X 6*^ X 6*^ X ... .
to n factors)
/'Qm+m + m+.. .. ton terni8\ y^ /'^c + c + c+.... ton termsA
— qM X n ^ ^c x n
— Qmn ^cn^ where c, m, and » are positive integers f a and 6
may be integral or fractional, positive or negative.
Similarly, (a"*6<^# joO" = o*«"6<^"cZ*« .... ;?♦•". Hence, in
general.
INVOLUTION. 63
To Baise a Monomial to any Power. Multiply the exponent
of each factor by tlie exponent of the required power y and take
the product of the resulting factors. Give to every even
poiver the positive siyyi, ami to every odd power the sign of
the monomial itsdf.
Notes : 1. Since, aw   1 X a"», the nth power of  a™ » ( 1 X a"*)"
— (— 1)* X «"•». Or we may write ± a'"»», for the nth power of  a»»», where
the positive or negative sign is to be prefixed, depeudiug upon the value of n
whether an even or odd integer, in being positive and integral.
2. Any power of a fraction is found by taking the required power of each of
Uliistrations.
(3x'»/)« = 3''<«a:«'<V'' =27x«y«
Exercise 20.
Write the results of the following :
1. (4aH*)2; (3a668)3; (2 a; V)^ (^a^V^c^f; (.1 a'»6)^
2. {1 a^V^f) (llaJ2c8(fiO)2. (3cx3/2«)8; {ZaH'^iff;
{5abcc^y^z^y.
3. (2a2)c2^7/*)8; (abcdxy^ (a^l^cf; (^^O*;
{SaPc^; {2aV^f.
4. (lXa<68c2^)n; (2a2"6"»)^ {Zxyzf\ (x^^^^^ ")•";
5. (2a''66«.)8. (^•)«. (a«)a. (ft ft ic 2)6; (m"7i— )•"•;
(2)8; («)«"; (ir.
54 ELEMENTS OF ALGEBRA.
6. n(2 a b'^ c n "i)*; n {n^ m^*"")"; m (m" a"^)";
7. 2 a( 3m7i3a;*)3; m( 3 aiojs^^e^*)*; a>^ (3 a^J^;
a(a«')«.
8. C2x^y\zhy\ (ra;V"^T; (3ain"c)";
(_ 3«''6't^iyA)6.
Affect the following with the exponent 7 ; that is, raise
each to the 7th power.
10. {x''y'^f{ahh^f;{^^a})xy\ [{x'^yf ', {T^mTf.
Write the nth. powers of:
11. mia'^ciyixyy; (a3 d)^ '' (xyf ; 3 (ab\c\d)
{a — xy.
12. a&c(a6T(^ + 2/ + ^T; a''{xy^zf''\xy'^f''\
28. It may be shown by actual multiplication that :
{a + hy =aH6H2a&;
(ahy =a^+h^2ab;
(a + b\cy =a2}62_,_c2+2a6 + 2ac + 26c;
(abcy =a^+b^\c^2ab2ac + 2bc;
{a\b + c\dy=a^+b^+c^\d^+2ab]2ac + 2ad + 2bc+2bd\2cd;
etc. etc. etc.
In each of the above products, observe that the square consists of :
I. The sum of the squares of the several terms of the given
expression.
II. Tvnce the algebraic product of the several terms taken
two and two.
INVOLUTION, 55
These laws hold good for the square of all expressions,
whatever be the number of terms. Hence, in general.
To Square any Polynomial Add together the squares of the
several tertns aiid twice the algebraic prodtict of every tvH)
terms.
Example 1. Square 3 a» — 4 x*.
Solution. The squares of the terms are 9 a' and 16 x^®. Twice
the algebraic product of the terms is — 24 a^ x*.
Therefore, (3 a»  4 x»)« = 9 a« + 16 x"  24 a«x».
Example 2. Square 2 x'  3 x*  1.
Solution. The squares of the terms are 4 x*, 9 x*, and 1. Twice
the algebraic product of the first term and each of the other two
terms gives the products — 12 x* and — 4 x*. Twice the product of
the second and third terms is 6 x^.
Therefore, (2x»3x2 l)a = 4x»+ 9x* I 1  12x»4x« + 6x«.
niuatrations.
(2 a"»  3 x»)« = (2a "•)2  ( 3 x— )«+ 2 (2 a"») X ( 3 x^")
= 4 a*" 4 9 r«"  12 a"» x"*.
(xVKy*+y'iy)'=(^V)"+(iartr*)*+(y^+(iyy
+ 2(xV) X ( ix"y«) + 2(r«y'')
X(§y'H2(xV)X(iy) + 2(ix"y«)
X(y«) + 2(ixir*)X(iy)+2(fy»)
x(Jy)
+ x«y*+tr«j/*+»§x"y + ix*y'»
56 ELEMENTS OF ALGEBRA.
Exercise 21.
Square, by inspection, the following :
1. x + 2; m + 5; n\7; a — 10; 2x \ Sy; a + Sb;
a — Sb; 2x — Sy.
2. X + 5y; 3x — 5y; 2a + ab; 5x — Sxy; 5abc — c;
xy^2y^; a™ + 3 6"".
S.2x + Sa^; xy + x'^; 3 a2 f 5 a^^; 1x;
1 — cy; m — 1; ab"^ — I ; J a"— .05.
4. 1 a 62 + 6ia;i; fffr^'^  a« _ 2 j»«;
icf 3/f + J; .0002a;'" + .005/.
5. I m^ n^p^ — ^mrf; xy + yzhxz; 2x^ ■] Sx — 1;
x^2x+l; x^ + 2x4.
6. 2a^—x + S; a^—5x—2; x^—2xy + y^; 4:n^+m^n—7i^;
x^  3x 4 2.
7. xy — 2n + 1; m — n — p — q; a^ — 2a^ \ 2x — 3;
1 + X } x'^ + x^; x+Sy+2a — b.
8. 2a^Sa^x + 3; x  2y  Zz + 2n\ wT ^ tT
9. ^a2bV\c',xfy + \a\b;la^x + l',
10. l^lx\x',la^\x\;\ar\a+\xy',
2 a;t + 5 a:i + 7.
11. '^x^2x^ + \x^x^; m^"f a;i«2/t*^'^3;
2i3i.
29. Any Power of a Binomial. It may be shown by
actual multiplication that:
INVOLUTION. 57
(a + 6)3 = a3 + 3 an + 3 a?;^ + 68.
(a  6)3 = a8  3 a26 + 3 a 6^  JS;
(a + 6)* = a* + 4a36 4 6 a^l^ + 4:ah^ ^ 6*;
(a  6)* = a*  4 a36 + 6 a2 62  4 a68 + 6*;
(,f + 6)'^ = a^+ 5a*6+ 10 a8 62 + 10 a263 4 5a6*+6«;
(,i 6)'^ = a65a*6+ 10a362 10a263+ 5a6*6fi;
and 80 on.
In each of the above products we obseive tlie following
laws:
I. The number of terms is one more than the exponent of
the binomial.
II. If both terms of the bin^omial are positive, all the terms
are positive,
III. If the second term of the binxrmial is negative, the
odd teims, in the product, are positive, and the even terms
negative.
IV. TTie first and the last terms of the product are respec
tively the first aiul the last terms of the binoinial raised to
the power to which tlie binomial is to be raised.
V. The exponent of tlie first tei^m of the binomial, in the
second term of the product, is one less than the exponent of
the binomial, and in each succeeding term it decreases by one.
The exponent of the second term of tlie binomial, in the
second term, of the product, is OTie, and in each succeeding
term it increases by one.
Thus, omitting coefficients,
(a + 6)« = a« + a^b + a*62 + aS^s ^ «2 j4 j^ ab^ ^ 6«
VI. The coefficient of the first and the last term is one,
that of the second term is the exponent of the binomial.
58 ELEMENTS OF ALGEBRA.
The coefficient of any term, multiplied by the exponent of
the first term of the binomial in that term, and divided by
the number of the term, will be the coefficient of the next term.
Notes : 1. The sum of the exponents in any term of the expansion is the
same, and is equal to the exponent of the binomial .
2. The coeflScients of terms equally distant from the first terra and the last
term of the expansion are equal. Thus, we may write out the coefficients of
the last half of the expansion from the first half.
If one or both of the terms of the binomial have more
than one literal factor, or a coefficient or exponent other
than 1, or if either of them is numerical, enclose it in
parentheses before applying the principles. Thus,
Example 1. Expand (2x'^5a^xy
Process.
(2 a:8  5 a2 a;)* = [ (2 a;3)  (5 a2 x) ]4
= {2x^y~4{2x^f{5a^x)\6{2xy{5a^x)^4{2x^){5a^xy
+ (5a^xy
= 24a;i2_4 X 2^x^ X 5a^x\6X 2^x^ X b^a^x^ 4 X 2a:8
X5^a^x^ + 5^a^x^
= 16a;i2_4X8a:«X 5a^x{6 X 4x«X 25a*a;24x 2x8
X 125a^x^+626a^x^
=z 16a;i2_160a2a;io + 600a4a:8_1000a6x«+625ft8a:*
Explanation. In the expansion the odd terms will be positive,
and the even terms negative. The first term is (2 x^y, and the fifth
or last is (5 a^xy. The exponent of (2 x^). is 4, and in each succeed
ing term it decreases by 1. The exponent of (5 a^x) is 1, and in each
succeeding term it increases by 1. The coefficient of the second term
is 4. For the second term we take the product of 4, (2 x^)^, and
(5 a^x). To find the coefficient of the third term, we multiply the
coefficient of the second term 4 by 3 (the exponent of (2 x^) in that
term), and divide the product by 2 (the number of the term), and
have 6. Hence, the third term is 6 (2 x^y (5 a^x)''^. The coefficient
of the fourth term is found by multiplying 6 (the coefficient of the
third term) by 2 (the exponent of (2 x^) in the third term), and
INVOLUTION. 59
dividing the product by 3 (the number of the term). Hence, the
fourth term is 4 (2 x*) (5 a* a:)*. Performing operations indicated, we
have the required result.
Example 2. Raise 1 — § x" to the fifth power.
Process.
(l§x")» = (l)»5(l)*(x«) + 10(l)»(§x»)«l0(l)«(fz»)» + 5(l)(fxn)*
= 1»5X l*xa*+10X l«xx210X l«X^x»« + 5X 1
Exercise 22.
Expand and simplify the following expressions :
1. (a  6)7; {a 4 x)^; {a^ ~ ac)*; (a^  4)^; (2 + a)*;
(«l)^ (1 aY; (2aSby.
2. (xh  3)* ; (axS x^f ; {x  3)^ (2 a^z + 3 62^)8;
(2ax\ Sbyy.
3. (« + 2)e; (a2)«; (2Ja)*; Cja36)*; (Ja + }6)*;
(a + 6)W.
4. (ai  2  a^y ; [(2: + y)H {x  yff\ (1 + a + a^)^
 (1  a + 2 a2)'f
5. (a + 2i)*(a26)*; (3  2a + a2)2  (2  a)*;
(3i + 5i)2 _ (2i  3i)2
Queries. How prove (— m)* = ± m", according to the value of
n, whether an even or odd integer ? How prove the method for
squaring any polynomial? How prove the laws for raising a bino
mial to any power?
60 ELEMENTS OF ALGEBRA.
CHAPTER VI.
ALGEBRAIC DIVISION.
30. Division is the inverse of multiplication, and is the
operation of finding the other factor, when a product and
one of its factors are given. The product is now called the
Dividend, the given factor is the Divisor, and the required
factor is the Quotient. Thus,
since a^ X a^ = a^ .*. a^ ^a^ = a^^
since a^Xa^^a^, .*. a^^a^ = a^\
since a^ X a* = a, .. a^a* — a^;
since a"*" X a" = a"*, .*. a"» r a" = a"»";
since «"»+" X a" = a"*, .*. a™ ^ a" = »*"+" ;
since Sa^ft* x 2a26 = 6a6^ .. Qah^ ^^aH = ^a^b^;
since 9a362 x 3a*65 = 27a6^ ... ^1 aU' ^Za^h^ = ^aH^;
since 5 a*6"~^ X 4a~*6^ = 20aH*, .•. 20a^h^^Aa~^h^ = baH~^ ;
etc. Hence, in general,
To Divide a Monomial by a Monomial. To the quotient of
the numerical coefficients annex the literal factors, each taken
'with an exponent obtained hy subtracting its exponent in the
divisor from its exponent in the dividend.
Illustrations.
a^h^c^m^^ a^b^c^m^ ap^b'^^c^'^m^'^ =abc*m.
63a26V5^ 7«36c'» = 9a2+3i2ic54 =()abc.
1^ 2A2 . «^ 5a2i2i ^a^b ,. ^ „
I5a^b^^6bc = — = (Art. 2).
2c 2 c ^ ^
ALGEBRAIC DIVISION. 61
Exercise 23.
Divide :
1. 3a362 by ab] I6a*i^ by 3aH^; 20a^}^c^ by 5a6V;
Smi by 5 m^.
2. tT^ by ?i~i^; a* by a*"^; a^js^n y^y ^352^2. ^m+i.
by a"* — ; 2'+' by 2''.
3. 15 at 6i a^ by 9 a2 6"! ic^. ^ ^i fti by f (A ji ;
21a*m2ic' by Tama:*.
4. 24a"j9'" by 3a>"; 36a'"mV^ by 9amyri*; «*'+y*
by a^/.
5. (x  y)6 by (a;  y^; (a  c)*+8 by {a  c)*i ;
fe*^^• by f 6/H^.
6. (6a3 62,; X iSftSJV) by (S^a^'c^ x 2a*c8); a*"' by
a*"; (2 77i7i")2* by (2mn«j^
31. Only a positive number, + a, when multiplied by + 6, can
give the positive product +a6. Therefore, +ab divided by +6 gives
the quotient + a.
Thus, since aX6 = + a6, .•. 4a6f + 6 = fa;
since aX6 = — a6, .*. — a6^ — 6 = a; ,
since — aX6 = — a6, .*. — a6^ + 6 = — a;
since — aX— 6 = 4a6, .*. { ab. — b = — a.
Hence, in general,
Law of Sig^. If the dividend and the divisor have the
same siguy the quotient is positive. If they have opposite
signSy the quotient is negative.
62 ELEMENTS OF ALGEBRA.
Example. Divide 12a"» by — 4 a".
Solution. Since there is a factor 4 in the divisor, there must be
a factor 3 in the quotient, in order to give a product of 12 in the divi
dend. Since there are m factors of a in the dividend, and n in the
divisor, there must be m — n factors of a in the quotient, in order to
give a product of a"» in the dividend. Hence, 120'" f  4 a*^ = — 3 a"*",
because only a negative number, — Sa"*"", when multiplied by —4a"
can give the positive product, 12 a*".
Illustrations.
 ISa^mHSf 3a2m*62   b a^^b^^m^'^ = ~5a^bm^.
 5 x^^y^z^ i 10 x^y^z^ = + ^ cc^os ^^85^63 ^ ^ ^x^y^z\
ia'{aby{x + y)^^ 4:a{ab)^{x + yy= a"^(a6) (xti/)"*".
Exercise 24.
Divide :
f. 6^ by 3^; 20aH^cJ by lOahc; 35a^^hy7a^;
laHc^ by 7aHc^.
2. 27ax* by 9^4. _a6^,6c6 by ^aHc'^; .^aH^^c^^
by fa^^iici*; 12^2"?/ 2 by f^j''?/.
3. Z\7n?n^3iP by 2i mi7i3^2;  5 m^j^iyio by
*lj2ga2m3x4?/; 3.2Jrts^?/S by 2.Qt2\a^xy\
4. .O^a^wiV^* t)y .0a2^2/2^3. _9.3m3«^2^«32/§c
by .3m3«+ix'*4 2/K
5. .^x'^ifhyix'y'^'^ J(a fe)3c8 by .6(a&)2ci0;
~ .3ai'"ft^ by .2 a" 6".
ALGEBRAIC DIVISION. 63
6.  .375 xi ?/U^^  y^^ by  i^ oc^ y (xi  i/l)l .
8mSi^r02y7 by 9 m ^ ,. " x^ y \
7. 1.2aiO(jcy)"r» hy .^a^{xyfz^'^\ m^n^xyY
(yzY by w 2«n2''(a; 7/) '•(?/ 2)i^.
Simplify the following, that is, perform the indicated
operations :
X ~.5a2''62«c2'.
9. {aH*^2ab) x 2a2fe2 x ( .Gaifti =  .3aUi).
10. (.3a'"6'*c''H .03a"'6c'') r 1 j «3'"63»c3''A:.
11. (4§«»6(ijc2fliaU3rf*)
X [6 a2c irf3 ^ ^ (84 ^8 j8 c f 7 a* b^ c^)].
12. (ic""^'* X aUi.i"") x (rtUa^^^^^i !6ir*'"7/t).
13. (Ua^H^c^^7aH*c^)
= (28 a^iV ^  4 a'^b^c").
14. (1.7«i6^cijc2M.l«2iia:8)x (a"'i*c6^al62c3).
32. Since (a •^ b) m = a m + b m, .. (am + bm) r m = a + b.
Since (a  b) m =■ am — bm, . • . (a m  6 w) f m = a  ft.
Since (iy2y«23x»ir') X 3xy»= 3x«y< + 6xi/*2 + 9x*^,
.. (3x»y*46xy»2 + 9x<y)T 3xi/» = 2r/2ya23x»y».
Hence, in general,
To Divide a Polynomial by a Monomial. Divide each term
of the dividend by the divisor, ami add the results.
64 ELEMENTS OF ALGEBRA.
Exercise 25.
Divide :
1. 2a?+ ^a^y ~ 8a^y^ by 2a^ 21 m^n^  1 m^n^
— 14 myi + 63 by 7 mn.
2. a^hc  a? b^ c2  a^ W' c^ + a^ b^ c^ by a^bc; 42 a^
— 1.1 a;2 + 28 ^ by .7 x,
3. 28a3 + 9a221a + 35 by 7a; 4: a^ b^  16 aH^
+ 4:aH^ by  4 aH^.
4. 66t262c3 _ 48^264^2 + 36^2^2^! _ 20abc^ by 4a&ca
5. 2.4 m27i2 — .8 mht^ — 2.4 m ?i2 + 4:m^n^ by .8 m n ;
icf — x^ y^ by ici.
6. "^a^^^ab^ac by 1.5a; .5 m57i2_ 3^3^* ^y
— 1.5 m^v?.
7.  72 o^ c2  48 a} ^10 + 32 ^2 c^ by 16 a^ c^; 3.6 n*
— 4.8 rtf by 4 n\ .
8. 11^2^3^ 3 a; 71 2.4 2/2 by .'^xy\ .09 7^1* 2.4 m^ 71
+ 4.8 «i5 by .03 wi*
9.  ft"* + 2a"'  3a" by  a^\ m"+i 7/i"+2 + 77i'*+3
— m"+* by m?.
10. 2.1 a 2:2 y"* 4 I.4a3;:c4^" 2.8a5^2^P by  ,1 axy"".
11. a'"53_^+ij2__^n2j by ab; 2a^a^jS.5a^x*
by 2.3^ a^x,
ALGEBRAIC DIVISION. 65
12. 2.25 a^x  .0625 aire  .375 a ex by .'67^ ax;
llxi33a;* by 11 x*.
13. 72 wt  60 mi ni + 12mi ni  6mT^ni by 24 mi.
14. 36 (x  yf  27 (xy)3 + lS{xy) by 9(a:y).
15. 123ry*2'S0sf''^^y^z+10SaP^y2f^^ by 6^y*^
16. m" (x — y)« — 7;ia(a; — t/)" by m* (a; — y)*.
17. {xhyY(xyy\(x+yy{xyy by (x^yy{xy)\
18. 2.5m2+1.6m7i+3.3m by .83m; a^iai^ift^+a ^
by aw.
33. It may be shown by actual multiplication that :
(TO+n+/>) (x+yfz) = ma:+TOy+m2 + nx + nT/ + nz+;)a;+/>y+/)2.
.♦. (nu:^my+'nu\nx\rny]rnz\px+py\pz)^{x + ]i + z) = m\n\p.
The division is performed as follows :
Separate the dividend into the three parts mx + myfmz,
n X 4 n y + n 2, and px + py\pz. The first term of the (quotient,
m, is found by dividing m x, the first term of the dividend, by x, the
first term of the <U visor ; multiplying the entire divisor by m will
produce the Jirst part of the dividend. The second term n of the
quotient is found by dividing the first term of the second part of the
dividend by the first term of the divisor ; multiplying the entire di
visor by n will produce the second part of the dividend. The third
term p of the quotient is found by dividing the first term of the third
part of the dividend by the first term h{ the divisor ; multiplying the
entire divisor by p will produce the third part of the dividend. The
work is conveniently arranged as follows ;
5
( ^MV£RS(Ty\
66
ELEMENTS OF ALGEBRA.
g
+
g
■v..^
w
M
f M
N
^i
fts
a,
^1
4
+
4
+
?=^
5>i
^
5ss
a^
a,
a,
a^
+
+
+
+
H
«
H
H
a.
^H
^i
a,
4
+
M
l^
?\i
$
8
s
+
+
+
?s»
5rj
?s^
S
S
S
+
+
+
«
H
H
s
s
S
4
N N
fc s
+ +
>i a>i
S g
+ +
« «
1. ^
N
^*"
+ ^
>. g
•+3
d
■^
d
quotie
O
'S
l
a
^^"^
"o
1
f
1
1
•^
^
o
•S
^,
^3
o
^
05
'^
^
^
^
'^u
c
^
si
i=l
d
a
^
rt
13
3
fi
rt
;•
Fi
o
o
d
O
w
Tl
•iH
'eS
^
%
_>
'>
'S
§
'S
OJ
^
O)
(B
2h
O)
^^
_t
J
c
"3
rt
d
O)
cj
<u
(u
%i
■^
V(
n:J
^_i
o
B
o
d
c3
o
u
S
M
u *
o
o
"73
o
:=i
3
rt
d
ij
g
^
n:)
o
o
Q
o
&4
;h
PU
S
fl^
m
PU
+
+
CO
I
I
^ I
I 7
I
CO
I
+ +
CM
H
H
1
CO
1
1
1
1— 1
4
H
(M
4
1i
CO
1
1
1*
1
1
1
1
«
1
1
CO
1
1
^
1
1
1
1
1
(?q
t
CO
CO
+
1
4
4
"«
TO
CO
CO
+
+
Tj
^
t,
(M
(M
1
i
1.
1.
1
1
. CM
1
1
fj"
o
.2
a>
f
^
J
1
m"
*>
^
d
0)
.^
d
•5
1
"I3
d
0)
d
■^3
%
^
^
+3
S
S
i
1
(N
CO
•11
1^
ptH
1
OQ
1
ALGEBRAIC DIVISION. 67
Explanation Dividing the first term of the dividend by the first
term of the divisor, we have a:*, the first term of the quotient. Now
as we are to find how many times x* — 3a:^j2z+lis contained in
the dividend, and have found that it is contained a:* times, we may
take X* time.s the divisor out of the dividend, and then proceed to find
how many times the divisor is contained in the iemainder of the divi
dend. Dividing the first term of the remainder by the first term of
the divisor, we have — 2 a:, the second term of the quotient. Simi
larly, we find the third term of the quotient. Hence, the quotient is
x«  2 X  2.
Notes: 1. Algebraic division is strictly analogous to *Mong division*' in
Arithmetic. The arrangement of the terms corresponding to the order of suc
cession of the thousands, hundreds, tens, units, etc., and the processes for both
are exactly the same.
2. It is convenient to arrange both dividend and divisor according to poioers
of the same letter ascending or descending.
3. It may happen the division cannot he exactly performed ; we then alge
biaicaUy add to the quotient the fraction whose numerator is the remainder,
and whose denominator is the divisor. Thus, if we divide x"^ — 2xy —i^ by
X — y, we shall obtain x — y in the quotient, and there vnll be a remainder
— 2ya. Hence, (xS  2xy  y^) r (x  y) = x  y  ~~. ■
X y
Example 2. Divide a* + 6' I c»  3 a 6 c by a + 6 f c.
Arranging acconling to the descending powers of a, we have:
Process. Dirisor. Diridend. Quotient
af6+c)a» 3a6c6»Hc»(a*<26ac
a* times the divisor, a»ta^fefa^
First remainder, aV)a*c 3a6c+6»4c»
— ab times the divisor, —aV* — ab^ — abc
Second remainder, a^(Hab^ 2a^c+6»+c»
— ac times the divisor, — q^c —ac^— ahc
Third remainder, ah^^ac^ abc{l^\c*
b^ times the divisor, ab^ ■ \b^\b^c
Fourth remainder, ac* abcb^c+c*
c* times the divi.sor, ac' ^bc^c*
Fifth and last remainder, abcb^cbc^
— be times the divisor, —abc—b^cbc^
To verify the work, multiply the quotient by the divisor.
68 ELEMENTS OF ALGEBRA.
Example 3. Divide i^ xy^ + \x^ + ^y^ by ^y + ^x.
Process.
^x + ^y)\x^ +^xy^+^y^{j^x^^xy + {y
Divisor X ^x\ ^x^ + ^x^y
First remainder, — i ^^2/ + ^ ^ 2/^ + i^y^
Divisor X — ^xy, — ^x'^y — ^ x y^
Second and last remainder, i ^ ^^ + iV 2/*
Divisor X \ y% i ^ J/" + i^ y^ 
Hence, in general,
To Divide a Polynomial by a Poljmomial. Divide tJie first
term of the dividend hy the first term of the divisor for the
first term of the quotient; multiply the entire divisor hy this
term, and subtract the product from the dividend. Divide
as before, and repeat the process until the work is completed.
Exercise 26.
Divide :
1. 14 x^ + 45 ^2/ + "^8 x^i/ + ^bxf +14.y^ hy 2x^
+ 5 xy + 7 y\
2. x'^ — 2x^y+2x^y^ — xy^ by x — y; a^ — 2ah^^h^
hj ab. ^
3. f5y^ + 9 2/6y^y+2 by y^  3 y + 2 ;
y^1 hy y1.
4. x"^ + xy + 2 xz—2y^+ 7 yz — 3 z^ hy x — y +3z;
c^ —b^ by a — h.
5. 2^2/+ 36?/ + 10fe;:c + 15 62 by 2/ + 5&; a6 + a^b
hy a \b.
6. .125ic32.25a;22^4. \Z,^ xy^ 27 y'^ by .^x3y,
7. ?/62/^2^ + 54^3a;2?/ by 2xy\ x^y^ by x + y.
ALGEBRAIC DIVISION. 69
8. a^t/^—a^ — y^+lhyxy + x\i/+l;4:i/ + 4:y
y8 by 3y + 2i/2 42.
9. s^+i^ — z^{3xyzhyx + y — z; x—y by x^ — yi.
10. 3^y^\2xi/^Z'a^z^{i/^z^ by xy + xz + yz;
x^ — y^ hy xi — yi.
11. 12x*26a^ySx^f+10xfSy^hySx^
2xy\f.
12. a^^'f^Zxy\hyx\y\) 5V ^ " iV ^
+ iV ^  S*I by J X  i.
13. 12 ;c« ?/9  14 x* 2/^ + 6 a;2 ^9 _ ^9 by 2x^y^f)
a^ — y^ by a;« — y^.
14. a^ 6  a ^ by a3 + 63 + a ^2 + ^2 J . ^ ^ ^4 _ ^
— x~^ by a; — x~^.
15. a^ + :r* y + a:^ 2/* + i^ ?/^ 4 a: ?/* + 7^5 by a;^ + 7/^ ;
al — 6f by a^ — fci
16. Ja3 + Y«^l25a + 2.25 by Ja + 3; .Ibx^y^
+ .048 a:^ by .2 a;^ ^ 5^,^
17. ata24at + Ga2ai by J4«i + 2; a:S_y5
by a; — ;/.
18. a;8 + 7^ + 23 4. 3 ^3y ^. 3 aj^J by a; + y + 2 ; .5 a;8
+ a.^+ .375 a: + 75 by J a; + 1.
19. x^\^y^^^^xyzhy 3^\^f^7?xz2xy
2yz.
20. ^g a:*  I a,3  J r^ + I a: + Jg^ by 1.5 a:^ _ 3, _ 
21. ofi — 1^ hy a^ + Qc^ y \ X y^ ■\ ij^ , a^ — 7^ by a^
+ a; y + 2/2.
70 ELEMENTS OF ALGEBRA.
22. 10 a^27aH + 34:a^I^lSah^8b^hj 5 a^6ab
 2R
23. 36x^+^i/+.254:X7j6x + ^y by Qx^.d.
24 ai2 + 2 aH^ + Z^^^ by «* + 2 ^2 ^2 + ^,4. ^6_2,6 ^y
a^2a^b + 2ah^h\
25. 2 ^^" — 6 iz2« y + 6 rr"2/2"  2 ?/3« ^y 2^ — 2/«; ^»
+ 2/3" by it" + ?/.
26. i6'2«— 7/2'" + 2 7/"*^' s2' by ^''+ ?/"• — ;3'; 32^22^
by 3^  2".
27. ifX^lilxi/^ by f^.752/; a:t"'3aji"'2/i''
+ 2 yh by a ^"^ — 2/i".
28. ?/2a;2m^2 7/22:'"+"+ 2 2/r a?"* + 22a;2n + 2ra?";2 + r2
by 3/ a:*" + 2; ^" + 7\
29. a;~i + 2x~^y~i + 2/"^ by a?~^ + y~^; x^ + ?/* by
a?2 + 22 a? 2/ + 2/2,
30. x~'^ — y'~'^\2y~^z~^ — z~'^ by a;~i + 2/~^~^~^j
a?* — 3 2/* by x — y.
34. There are special methods for finding the quotient
of binomials, hy inspection, which are of importance on ac
count of their frequent occurrence in algebraic operations.
Thus,
It may be shown by actual division that :
a—b ^a—b
^^—^ = a^+as 6+a2 h^+ab^+b^ ; ^"—^ = a^ + a'^b + a%'^ + 02^8 +ab^+b^)
a—b a—b
and so on. Hence, in general, it will be found that,
ALGEBRAIC DIVISION. 71
The difference of any two equal powers of two numbers is
divisible by the difference of the numbers.
In each of the above quotients we observe the following
laws :
I. The number of terms is equal to the exponent of the
powers.
II. Hie signs are all positive.
III. The exponent of a in the first term is one less than
the exponent of a in the first term of the dividend, and in
each succeeding term it decreases by one {in the last term its
exponent is 0, or a. disappears).
The exponent of b in the second term is one, and in each
succeeding term it increases by one {in the last term its expo
nent is one less than the exponent of b in the dividend).
IV. The first term is found by dividing the first ter^n of
the dividend by tJie first term of the divisor.
V. To find each succeeding term, divide the preceding
term by the first term of the divisor, and multiply the restUt
by the second term of the divisor regardless of sign.
Example. Divide 1 — w* by 1 — n.
Solution. Dividing 1, the first term of the dividend, by 1, the
first term of the divisor, we get 1 for the first term of the quotient
Now divide the first term of the quotient by the first term of the
divisor, and multiply the result by n, the second term of the divisor
(regardless of sign), for the second term, n, of the quotient. Dividing
the second term of the quotient by the first term of the divisor, and
multiplying the result by n, we have n^ for the third term of the quo
tient. Similarly, we find n*, and n* for the fourth and ffih terms,
respectively. .*. (1 — n*) r (1 — n3 = 1  n f n* + n* + n*.
72 ELEMENTS OF ALGEBRA.
Exercise 27.
Divide by inspection :
1. m^ — n^ by m — n\ a^ m^ — h^n^ by am — hn\
m^n^ —1 by mn — 1.
2. l—m^n^a^hjl—mnxjix yj*— {x zj hy xy — xz;
1 — a''b'' x'^ hj 1 — ahx.
In order to apply this principle the terms of the divi
dend must be the same powers of the respective terms of
the divisor. It is not necessary that the exponents of the
terms of the divisor be 1, nor that they be the same, nor
that the exponents of the terms of the dividend be the
same. Thus,
Example Divide x^^ — y^^ by x^ — y*.
iSolution Dividing x^^ by x^, we have x^ for the ^rst term in the
quotient. Now divide x^ hj x^ and multiply the result by y*, for the
second term, x^ y*, in the quotient. In like manner we find x^y% and
y^^ for the third and fourth teims of the quotient.
. •. (a;i2  i/16) i (x3  2/4) = a;9 + ^6 ?y4 + a:3 2/8 + ^12
So in general x"* — y"^ divides 2:"" — ;v""* {n being any
positive integer), since the dividend is the difference be
tween the nVa powers of the terms of the divisor.
3. a^  W by a^l^ x^^  f^ by x^  y'^ \ x^^  y^'^
by a::^ — y'^.
4 ai5  &30 ]3y a3  &6 . x^m _ ^35n ^^ ^n _ ^n . 2io«
 a^"" by 22"  x"^.
We may easily apply these principles to examples con
taining coefficients as well as exponents; also to those
involving fractional or negative exponents. Thus,
ALGEBRAIC DIVISION. 73
Example. Divide 81 a^^  16 A** by 3 a«  2 h\
Solution. Dividing 81 a" by 3 a«, we have 27 a» for the frst
term of the quotient. Now divide 27 a* by 3 a* and multiply the
result by 2 6*, for the second term, 18 a® 6*, in the quotient. Simi
larly, we find 12 a' 6^*, and 8 h^^ for the <At>d and fourth terms in the
quotient.
.. (81a"16634)^(3a»26«) = 27a»+18a«6«+12a«6»2+86".
If a and h are coefficients, a^a^** ~ 6"^*"" is divisible by
ax' — hif^f since the dividend is the difference between
^1 1
the Tith powers of ax* and ////'". In general, a? "•— y *"
na n$
divides x '^ — y "" (n being any positive integer), since
a
the latter is the difference between the ?ith powers of a; •*
and y '.
5. 64ai2_27?i» by 4a^Sn^; 16 2^ y^^"*  ^ m^z^*
by 4a:8y6*_ j^^4 2io»
6. «8 a^S"  68 2/3« by a2aJJi' _ ^^ ^^^ . 32 ar^o  243 y^
by 2 0^2  3 3/3.
7. x~^ — y~i by a;~i — y~i ; 3^ — y^ by a;^ — yi ;
a xi — 6i y by ai x^ — fci yi
35. It may be shown by actual division that :
ai—l)i a*h*
— rj = a — b; — —r = a*—a^b{ab^—b*; and so on.
a+o a + o
Hence, in general, it will be found that,
The difference of any two equal even powers of two num
bers is divisible by the sum of the numbers.
In each of the above quotients we observe the laws are
the same as in I. and III., Art. 34 ; also,
74 ELEMENTS OP ALGEBRA.
VI. The signs are alternately + and — .
Hence, the principle may be applied to different classes
of examples as in Art. 34. Thus, in general,
If a and l are coefficients, ax^ {■ h'lf divides a" x"*^
— If y^"^ (n being any even and positive integer ; also m and
p may be integral, fractional, or negative), since the divi
dend is the difference between the nth. powers of a x^ and
by.
Note, 'rtie difference of the squai'es of two numbers is always divisible by
the sum and also by the difference of the numbers. Thus, 06 — 68 jg divisible
by aS ± 64. jn general, a^n — IZm [§ divisible by a« ± **" when n and m are
integral. This is the converse of Art. 26.
Exercise 28.
Divide by inspection :
1. 625 a^x^ — 81 ra^n^ by 5ax+ 3 mn; a^ — h^ hy
x:^ + b^; x^ — 1 by x + 1.
2. x^  yi by x^ + y^ ; 256 ir*  10000 by 4£c + 10^,
3 ^im^yin by a;"* + 2/" ; tV ^^  ^^^^ 2^^ by J rr^ + .22/1 ;
^10 _ ^,10 ]^,y a, + 6.
4. 729 ai2 _ 64 ^,18 ^y S a^ + 2 b^ ; a^ x^""  ¥y^"* by
a ^^ + &2 y2 m
5. a""^ — x'"^ by a~^ + a;"^; a^ x~^ — b^y~^ by
a a:~3 + h's y~^ •
6. xh  yi"" by 2;i*^ + t/t^"*; 81ai"a;  Jq &f«^f'»
by 3 aT5'*a;4 +  b^y'^'^.
ALGEBRAIC DIVISION. 75
36. It may be shown by actual division that :
j = a^—ab + b^', j = a*—a*b+a^b^—ab^^b*; and so on.
a\b a + b
Hence, in general, it will be found that,
The sum of any two equal odd powers of two numbers is
divisible by the sum of the numbers.
In each of the above quotients we observe that the laws
are the same as in Art. 35.
Hence, the principle may be applied to all the different
classes of examples as in Art. 34. Thus, in general,
If a and b are coefficients, aaf^hy"^ divides a" a:"'
_j_ i^ynm ^^^ beiug ani/ odd and positive integer, also m and
p are integral, fractional, or negative), since the dividend
is the sum of the nth powers of aaf and by^.
Exercise 29.
Divide by insj^ection :
1. x" ■\ If by X \ y \ x~^ + y~^ by a;"^ + 3rM
1024 a:« + 243 y« by 4 a; + 3 y.
2. 128 x^^ + 2187 y" by 2 2:3 + 3 y2 . 243 x^^ + 32 ^o
by 3 a:8 4. 2 y2.
3. a;!*" + 2/21*" by 3^^ ■¥ y^^  A:2i« 4 7^86 ^y l^^^m^''',
a" + 6" by a + 6.
4. wi w + a; y by wi ?ii + xi yi ; x^ + ?/V by x^ \~ yi\
^\\ y'^ by a;i + yK
5. ax\^b^yhya\x\^biiy\', (# f ( J) by (f )i + (f )i ;
a^ + 6^0 by a + 62.
Note. Since a« and Ifi are odd powers of a^ and h^, n^ + 6« is divisible by
cfl + 62. aW and ft" are the 5th powers of a^ and 62, a 10 + 6W is divisible by
a* + 6^. Also, a» and 6* are the third powers of a« and 6«. Therefore, a« f 6»
is divisible by a* + b^.
76 ELEMENTS OF ALGEBRA.
6. a^ + 612 by a* + &^ 2:6 + 1 ]3y x^ +,1 ', x^^ + 1 by
a;* + 1 ; a27 + ^27 by a^ + h^.
7. a;io + 2/10 by 2;2 + 2/2 ; a;!^ + 2/^^ by a^ + y^ ; 64: + x^
by 22 4 2;2
8. 64 x^ + 729 f by 22 a;2 + 9 ^^2 . ^lo + __l^ by x^ + ( J)2;
^24 4. 524 by «§ + &8 ■
9. ai8 + Z,i8 by a^ + b^ and by a^+P; j^^ x^ + J^ f
by i^^ + i2/^
10. ^36 + J36 by ai2 + ^,12 and by a^ + J*; 729 a;^ + 1
by 9 a:2 + 1.
11. a;*2 + 2/42 by x^ + ^6 and by x^^ + 2/^*. Query. Is
it divisible by 2;2 + / ? Why ? a^^ + &18 by a^ + h^ ',
^27 + J21 by a9 + &' ; a^^ + 515"' by a^ + J^.
12. rr^ + 2/^* by o^^^ + ?/i^ and by x^ + /. Query. Is
it divisible by 2^2 + 2/^ ? Why ? a^ + &12 by a^ _^ &* ;
aH9 + m27n36 by a^h^ + 77i9 7ii2; ^15^25 _^ m^o^iio by
a^ 6^ + m^ 7i2
Find an exact divisor and the quotient for each of the
following, by inspection :
13. 8 + a^; ^6  6^ 8  ^; a:*  81; a^^  h^^;
81 ai2  16 h^; a"^ 625 ; a^  h^
14. aj20  2/15; ^,^5 + ^5. ^12 _ ^12 . ^6 _ 1 . ^12 _ 2,12.
a^a^^h^fp a^'62) 16 «*  81.
15. Z2w>  h^; 81 «8  16 b^; 1  y'^ ; a^ x^ + 1000;
a* a;* — 1 ; a^ + m^a:^ ; 2^2 7/2 — 81 (x2
16. 32 ^10  243 2>i5. cIo.tIO'*  a^^x^^; a^" + ftQ''^
a^x^ + b^y^""; c^ x^^ + 6' 2/^".
ALGEBRAIC DIVISION. 77
17. aj^" + ye*; 8a^y3 4. 729; fti2 yi2j» _ Ji:^yi2».
cS^^Sp _ j8^«^ 2:4 _ 1296; ^^S"  ftio*
18. 128.^21 + 2187^^*; 256a;i281 //«; a6»65* + a:S'y^';
1 + 128 a:!*; ««'» d^^. ^jii^il.
19. x^yi_aiyi; ai^S+l; i^a^x^'^^^^h'^f^
g^ xt". 00032 y^".
20. ii c" + .002432;2/i; 256 aa;t~  .0081 Jf*;
Queries. How divide a monomial by a monomial ? Prove it.
How prove the method lor dividing a polynomial by a polynomial \
In Art. 35, the sign of the last term ot" the quotient is — , while in
Art. 36, the sign ot* the last term of the quotient is 4. Why is this ?
What is the product of a* and a * ? Prove it. What of a* and or* I
Prove it.
Miscellaneous Exercise 30.
Divide :
1. a3 63+ T^a2 62 + ^1 by Jafea + JJi; ari + yi
by x~^ + ir^
2. a6+ 5aHi+10ft362f i0a2 63 + bcrH'^
+ 66 by ai + b\
3. 2 a2  a^  2 rt + 1 by 1  a^ ; x  ij hy x\  iji.
4. (a — 6  c)"  (a  6  c)" "•  (rt — 6 — c)"" by
(a hcyr
5. 2 2^3 + 2 y3 + 2z^ 6zyzhy (x  yf \ (y  zf
+ (z2:)2; (2^ff by {x' + xy + ff.
6. {x^2yzf^f7? by ^^24^2; (a:+2y)8 + (y3«)3
by ar + 3 (y  z).
78 ELEMENTS OF ALGEBRA.
7. x^ — x^i/ \ xy^ — 2 xi y^ + y^ by xi — xy^ \ x^y
yl
8. 2 :r3« _ 6 ,jc2nyn _j_ e ^n^2n _ 2 fn ^^ ^n _ ^n^
by :r"'  3 a;'""^ /  6 a:'"^ ?/2«.
10. a3"  3 a2« 5« + 3 a« &2« _ 53^ by ^« _ j« . 3 ^o:
 8 a^ + 5 ^3^  3 a3^ by 5 a^  3 ^^
11. 6 «'« + '  23 a^" + ^ + 18 a^  a^"^  3 a'"~'
4. 4a3«3  a""' by 2 a^''^^  5 a^"  2a2'*^ + a'''.
12. 4 at  8 ai  5 + 10 ^^ + 3 a"! by 2 ai2  aiV
 3 h~l
13. 6:r^+35:2r" + '6 2r^ + ^+19af21r^^ +4,r^2
by 2 x^ + x^ — 4:X.
14. 6m*"*2 + m*"+^  22 m^"" + 19 7?t*"^
15. 6aj^ + "+'+2f^"^^9af + "+lla;^ + "'6;zf + "2
^^ + n3 by 2 2^'^^'+ 3a;" + ^a;".
16. a""* — ct" fe(''i)'» — a<"*i>" &"* + 5"*" by a"  &**.
Find an exact divisor and the quotient of the following,
by inspection :
17. 8a^+l; 16  81 a^ ; 64 a^  8 &3 . ^34. iqOO ;
^6  64 ; m^ 71^; 1  8 ?/3 ; a^ 6^  1.
19. 8 ;x^6  27 ?/^ 64 ai2 _ 27 ^9; 243 a^ + 32;
cSjc^'*  ftS^Sm . 1^ ^4n _ 0016 hh'^', 29" al2» + 36n
EVOLUTION. 79
CHAPTER VIL
EVOLUTION.
37. Evolution is the operation of finding one of the
equal factors of a number or expression. Evolution is the
inverse of involution.
By Art. 27, (2 a)» = 4 a^; (2 «)» = 8 a* ; (2 a)* =16 a*; etc.
2 a is called the secoml or sqiuire root of 4 o^ l)ecause it is one
of the two ec[ual factors ot 4 a'^; it is the thiid or cube root of 8 a*
because it is one of the three equal factors of 8a^; etc. Hence, in
general,
A Root is one of the equal factors of the number or
expression.
Roots are indicated by means of fractional exponents,
the denominators of which show the root to be taken.
Thus, (a)* means the second or square root of a ; (a)* means the
third or cube root of a; (a*)* means the sixth root of a^. In general,
(a*")* means the nth root of a"*.
Roots are also indicated by means of the root sign, or
radical sign, ^.
Thus, \/a means the square root of a ; ^a means the cube root
of a ; v^ means the nth root of a"».
The Index is the number written ia the opening of the
radical sign to show wliat root is sought, and corresponds
to the denominator of the fractional exponent. When no
index is written, the square root la understood.
80 ELEMENTS OF ALGEBRA.
« — 1
Note, ya or a» is defined, when n is a positive integer, as one of the n
equal factors of a ; so that if Va be taken n times as a factor, the resulting
product is a ; that is, ( ya)'^ or i^a" j" = a.
,mn \mn ( ^\mn,
Similarly, ( \/a) or Va*""/ = a,
38. The sign, ± or =p, is sometimes used and is called the double
sign; it indicates that we may take either the sign + or the sign — .
Thus, a ± 6 is read a plus or minus b.
By Art. 27, (+a)4 = a^ (a)4 = a4; (+a)5 = a5; (ay = aK
Therefore, (a*)^ = ±a; (+ a^)^ = a; (— a^)^ = — a. Hence, in
general,
Hven roots of any nu7nber are either positive or negative.
Odd roots of a nwniber kave the same sign as the number
itself.
Since no even power of a number can be negative, it
follows that,
An even root of a negative number is impossible.
Such roots can only be indicated, and are called imaginary. Thus,
(— a^)^y Y^— 6, Y^— 1, and ^— a^, are imaginary.
Example 1. Find the square root of 9 a^b^c^.
Solution. Since, to square a monomial, we multiply the expo
.nent of each factor by 2, to extract the square root we must divide
the exponent of each factor by 2. The two equal factors of 9 are
3 X 3, or 32. Dividing the exponent of each factor by 2, we have
3 a^ b'^ c. Since the even root of a positive number is either positive
or negative, the sign of the root is either plus or minus.
.. y/9aH^ = ±3a^b^c.
Example 2. Find the fifth root of  32 a^^ x"*.
Solution. Since, to raise a monomial to the fifth power, we mul
tiply the exponent of each factor by 5, to extract the fifth root we
must divide the exponent of each factor by 5. The equal factors of
32 are 2 X 2 X 2 X 2 X 2, or 2^. Dividing the exponent of each
EVOLUTION. 81
factor by 6, we have 2 a* x*. Since the odd roots of a number have
the same sign a.s the number itself, the sign of the rout is minus.
.♦. J^— 32 a*® x*» = — 2 a* a*. Hence, in general,
To find any Boot of a Monomial Resolve ike numerical
coejffkient into its prime factors, each factor being written
with its highest exponent, divide the exponent of each factor
by the index of the required root, and take the ^product of
the resulting factors. Give to every even root of a positive
expression tlie sign ± , and to every odd root of any expres
sion the sign of the expression itself
Hote. Any root of a fraction is found by taking the required root of each
V27
of ito terms. Thus, .Vl _ J^ = 2 in general, t/ = ^
Exercise 31.
Find the value of the following expressions :
1. V25^; ^%aH^a^', y/12baH^', v^81ai«W
2. (343a*56)i; {\{7^ifz^)^; {x^^y^^)^ v'Sl^V^.
3. v^iw. (I21a:i2y2)i. V25 aH"^, (16a8&8)i
4. ( 243 a^* Jio«)i. (_ 54 ,,,3 ^6 ^^J. (^w ^80)tV.
5. (32ai0y5)i; V^2W^r^ , (625 a^ 6i« c*)i.
6. (512an8ci5rf8)i; V64a«&"*; y/m ; ^/ 32 ai^.
8. V'i^^; (2"a2«54n2^)i; V8l2:5iy2m+4. >^_ 8 a;8*«ye+».
A
82 * ELEMENTS OF ALGEBRA.
10. \/l6x^''y^''z^; ( iV^ ^^ ^ ?i" ^M ; V^^^Sm^m
Simplify :
11. Sj^a^hUl + iij a^ hh d)^  (f i a* ol c^)i
— V ^^ <^^« c 4.
V 50
Express the nth roots of:
12. 3x7x4; 52:"2/2^ 3a^&3. (air)3; (^^z)"; ^'"y";
Express by means of exponents :
13. \/JWc^; 7a(xyT; V^^^; V^(^ + 2/)~
Queries. If n and j9 in the last two parts of Ex. 13 are integral,
what signs should the roots have ? Why ? When should the first
two roots have the double sign ?
39. By Art. 28, (a + hy = a^ + 2ah \ b^.
Therefore, (a^ + 2 a 6 + 62)i = a^b.
By observing the manner in which a + b may be obtained from
a^ + 2ab + b^, we shall be led to a general method for finding the
square root of any polynomial.
Process. a"^ + 2ab h b^(a + b
First term of the root squared, a^
First remainder, 2 ab + b^
Trial divisor, 2 a
Complete divisor, 2a + b
Complete divisor X b, 2ab + b^
Explanation. The square root of the first term is a, which is
the first term of the required root. Subtracting its square from the
given expression, the remainder is 2ab + b^, or b times 2 a \ b.
EVOLUTION. 83
Since the first terra of the ieinainder is twice the product of the fiist
and last terras of the root, and we have found the first terra ; there
fore, divide 2 a6 by twice the fiist terra of the root already found, or
2 a. 'I'he result will be the second terra b of the required root.
Adding 6 to the trial divisor gives the complete divisor, 2 a + 6.
Multiplying by 6 and subtracting, there is no remainder.
By Art 28, (a\b+cy = a^+2ab{b^+2ac + 2bc + c^.
Therefore, (a^{2ab^b^\2ac^2bc + c^^ = a + b+c.
Process. a^\2ab\b^+2ac^2bc + c'^{a\b + c
First terra of root squared, a^
First remainder, 2ab + b^{2ac + 2bc+c^
First trial divisor, 2 a
First complete divisor, 2a\b
First complete divisor X 6, 2ab{b^
Second remainder, 2ac\2bc^c*
Second trial divisor, 2a26 I
Second complete divisor, 2a\2bic \
Second complete divisor X c, 2ac+2bc\c^
Xtxplanation. Proceeding as before, the first two terms of the
root are found to be a + 6. To find the last terra of the root, take
twice the terms of the root already found for the second trial divisor.
Dividing 2 a c by the first terra, the result c will be the third term of
the required root. Adding this to the trial divisor, gives the entire
divi.<*or. Multiplying by c and subtracting there is no remainder.
We have actually squared the root and subtracted the square from
the given expression. Hence, in general,
To find the Square Root of any Polynomial Arrange the
terms according to the powers of one letter. Find the square root
of the first term. This will be the first term of the required root.
Subtract its sciuare from the given expression. Divide the first term
of the remainder by twice the root already found. The quotient
will be the next term of the root. Add the quotient to the divisor.
Multiply the complete divisor by this terra of the root, and subtract
the product frora the remainder. For the next trial divisor, take
two times the terms of the root already found. Continue in this
manner until there i* no remainder.
84
ELEMENTS OF ALGEBRA.
Example. Find the square root of 4 a — 10 a^ + a^ + 4 a^ + 1.
Arranging according to the ascending powers of a, we have,
[_2a2a3
Process.
First term of root squared,
First remainder,
First trial divisor, 2
First complete divisoi' , 2+2a
2 a times first complete divisor,
Second remainder,
Second trial divisor, 2+4 a
Second complete divisor, 2+4 a— 2 a^
1+4 a 10a3+4a5+a6(i42a
_1
4a10a3+4a5+a6
4a^+4q
4a'^10a3+4a5+a«
— 2a^ times second complete divisor,
Third remainder,
Third trial divisor, 2+4 a— 4 a^
Third complete divisor, 2+4 a— 4 a^—cfi
4a28a8+4a^
2a84a4+4a6(a«
a* times third complete divisor.
~2a34a4+4a5+a«
Note. The student should notice that the sum of the several subtrahends
is the square of the root, and that he has actually squared the root and sub
tracted the square from the given expression.
Exercise 32.
Find the square roots of:
1. 2^4^3+ (]y2_4y4.i. 9^4 12 a3_ 2^2+4^ + 1.
2. 4a6_ l2a^l~ 11 a^^ly^ + b^ a^ h^  17 aH^ 70 ab^
+ 49 &6.
3. x^  12 a;5 + 60 x^  160 a^ + 240 x'^  192 a; + 64.
4. 8 a + 4 + a^ + 4 ^3 + 8 a2
5. 9 + a;6 4 30 :r  4 a^ + 13 x^ + Ux^  Ux^,
6. 6a62c_4a2&c + ^252 + 4ft2c2 + 9 52^2  12 a.&c2
7. 49 a^  28 a^  17 a^ + 6 « 4 f.
EVOLUTION. 85
8. 4x^ + 9 y^ + 25 cc^^ 12x1/ SO ay 20 ax,
9. VL^ — 6 a 771^ + 15 a2 7n* — 20 a^ vi^ + 15 a* m^ — 6a^7n
10. 1 2 a + 3 a2  4 a3 4 5 rt*  4 a5+ 3 a«— 2 a^ + a^.
11. 9 vi^ — G 7« n + 30mx+6my\7i^—i0nx—2 ny
+ 25r»+ lUa;^+ //2.
12. 7^\l5x^{/^^ 15aj*//2 4. ,/6_ ^y _ 20xV <^^"^y
13. 49x^/ 24x^/3 30 ar8y + 25 a^+ 163^.
14. a;6  G 2:^ + 172:4  34 2:3 _^ 40 a:2 _ 40 2: + 35.
15. 4  IG «t + IG a^ + 12 rf  24 at ?>i + 9 h.
10. ^a:4_^y^_^,,2,,2_,^y^^,^4. 2:4«_ 0a;3n+ 5^*
4 12 a;" + 4.
17. 25 ^1 + 1 G  30 X 24:xh + 49 .rf
18. 9x2+ 122;V^6a; + 42^4ar^7/^ + ^.
40. Since the square root of an expression is either + or — , the
square root o[ a^ \ '2 a h \ h^ is either a + h or — a — h. In the
process of finding the sfjuare root of a'* + 2 a 6 + />', we herein by tak
ing the square root of a*, and this is either + a or — a. If we take
— a, and continue the work as in Art. 39, we get for the root ^a^h.
Also, the square root of a' — 2 a 6 4 />^ is either a — h ov — a { b.
This is true for every even root. Hence, the signs of all the terms oj
an even root may he changed^ and the number will still be the root oj the
same expression. Thus, last process Art. 39, if — 1 he taken for
the square root of 1 we shall arrive at the result — 1 — 2 a + 2 a'* 4 a*.
41. Square Root of Numerical Numbers. The method
for extracting the square root of arithmetical numbers is
based upon the algebraic method.
86 ELEMENTS OF ALGEBRA.
Since the square root of 100 is 10, of 10000 is 100, etc., it fol
lows that the integral part of the square root of numbers less than 100
has one figure, of numbers between 100 and 10000 two figures, and so
on. Hence,
If a point he placed over every second figure in any numher, begin
ning with units' place, the numher oj points ivill show the numher of
figures in the square root.
Thus, the square root of 324947 has three figures ; the square root
of 441 has two figures. If the given number contains decimals, the
number of decimal places in the square root will be one half as many
as in the given number itself. Thus, if 2.39 be the square root, the
number will be 5.Vi2i; if .239 be the root, the number will be
6.057121 ; if 10.321 be the root, the number will be 106.523041.
Hence,
The numher of points to the left of the decimal point will shorn the
numher of integral places in the root, and the numher of points to the
right will show the numher of decimal places.
Example 1. Find the square root of 45796.
a +&+c = 214
Process. 45796(200+10+4 = 214
The square of a or 200,  40000
First remainder, 5796
First trial divisor, 2 a, or 400 I
First complete divisor, 2alh, or 410 [
First complete divisor X &, or 10, 4100
Second remainder, 1696
Second trial divisor, 2a+2&, or 420 i
Second complete divisor , 2a+2&+c, or 424 
Second complete divisor X c, or 4, 1696
Explanation. There will be three figures in the root. Let
a ^h \ c denote the root, a being the value of the number in the
hundreds' place, h of that in the tens' place, and c the number in the
units' place.
Then a must be the greatest multiple of 100 whose square is less
than 45796, this is 200. Subtract a^, or the square of 200 from the
given number. Dividing the first remainder by 2 a, or 400, gives 10
EVOLUTION.
87
for the value of b. Add this to 400, multiply the result by 10 and
subtract. Dividing the second remainder by 2 a f 2 6, or 420, gives
4 for the value of c. Adding this to 420, multiplying and subtract
ing, there is no remainder. Hence, 214 is the required root; because
we have actually squared it and subtracted this square from the
given number and found no remainder. The student should observe
that the .sum of the several subtrahends is the square of the root.
Example 2. Find the square root of 17.3 to lour decimal places.
Process.
S<iuare of 4,
First remainder,
First trial divisor, 8
First complete divisor, 81
First complete divisor multiplied by 1,
Second remainder.
Second trial divisor, 82
Second complete divisor, 825
17.360<KK)06(4.1693...,
16
130
81
4900
Second complete divisor nmltiplied by 5,
Third remainder,
Third trial divisor, 830
Third complete divisor, 8309
Third complete divisor multiplied by 9,
Fourth remainder.
Fourth trial divisor, 8318 I
Fourth complete divisor, 83183 
Fourth complete divisor multiplied by 3,
Fifth remainder,
4125
77500
74781
271900
249549
22351
Let the student formulate a method for arithmetical square root
from what has been demonstrated.
NotM ; 1. If the trial divisor is not contained in the remainder, annex to
the root, also to the divisor, then annex the next period and divide.
2. Should it be found that after completing the trial divisor, it gives a pro
duct greater than the remainder, the quotient is too large, and a less quotient
must be taken.
3. Tf the last remainder is not a perfect square, annex periods of ciphers and
procee<l as before.
4. The square root of a fraction may be found by taking the square root of
its terms, or by first reducing it to a decimal.
88 ELEMENTS OF ALGEBRA.
Exercise 33.
Find the square roots of :
1. 33124; 41.2164; Jf  ; ^%^^^; .099225; 1.170724.
2. .30858025; 5687573056; 943042681.
Find the square root to four decimal places of:
3. .081; .9; .001; .144; if; .00028561; 3.25; 20.911.
42. By Art. 29, (a + by = a^ + 3 a^ b + 3 ab^ + bK
Therefore, (a^ + 3 a^ 6 + 3 a &2 + b^)l = a + b.
By observing the manner in which a \b may be obtained from
a^ + 3 a^ft + 3 a &2 I 6"^, we shall be led to a general method for find
ing the cube root of any compound expression.
Process. a^+2a%yMb'^\b^ {a+b
First term of the root cubed, o^
First remainder, 3a%+Zab'^{b^
Trial divisor, or 3 times the square of a, Za^
3 times the product of a and b, 2ab
Second term of the root squared, 6^
Complete divisor, Za^+Zab+b'^
Complete divisor X &, Za%+Zab'^^b^
Explanation. The cube root of the first term is a, which is the
first term of the required root. Subtracting its cube from the given
expression, the remainder hZaH + Zab'^+b'^, or b times Za^ + Zab+b^
Since the first term of the remainder is three times the product of the
square of the first term of the root multiplied by the last term, divide
Za^b by three times the square of the first term of the root already
found. The result will be the second term 6 of the required root.
Adding to the trial divisor three times the product of the first and
second terms of the root, and the square of the second term, gives the
complete divisor, ov Zw^^ Zab + b^. Multiplying by b and subtract
ing, there is no remainder.
Since the cube of a+b\c is a^\Za'^b^Zab'^\h^+Za^c + Qabc
f3&2c + 3ac2+36cHc8, the cube root of a8 + 30^6  3 a 62+63+ 3 a^c
+ 6a6c + 362c+3acH36c2+c« is a + b + c.
EVOLUTION.
89
i
CO
4
^
^j
o
eo
cS
j;
w
^
1
t
s
5

^
^
^
t
^
1
•o
e^
«
$
#
c o
J
cS
<Sc^'"
^5 ^
o
I
« ^ X
^
iz
•5 S ,2
O C ;:2
'tis k.)
I
Ph C
■2 5
^ r ^ « ^ •>
g ^ 3 B .5 ^
"^ q O) 4> eS ^_^
•2 iJ s S S is Ts
£ 3 8 8 2 '^ §
£ I ^ ^ § §^
3,3.
s a
8 8
o p
90 ELEMENTS OF ALGEBRA.
Explanation. Proceeding as before, the first two terms of the
root are found to be a + h. To find the last term of the root, take
three times the square of the terms of the root already found for the
second trial divisor, and divide Za^c by the first term. The result
will be the third term of the required root. Adding to the second
trial divisor three times the product of a + & and c, and the square of
c, gives the second complete divisor. Multiplying by c and subtract
ing, there is no remainder. Observe that the sum of the several sub
trahends is the cube of the root, and that we have actually cubed the
root and subtracted the cube from the given expression. Hence, in
general,
To find the Cube Root of any Polynomial. Arrange the terms
according to the powers of one letter. Find the cube root of the first
term. This will be the first term of the required root. Subtract
its cube from the given expression. Divide the first term of the
remainder by three times the square of the root already found. The
quotient will be the next term of the root. Add to the trial divisor
three times the product of the first and second terms of the root, and
the square of the second term. Multiply the complete divisor by
this term of the root, and subtract the product from the remainder.
For the next trial divisor, take three times the square of the root
already found. Continue in this manner until there is no remainder
or an approximate root found.
A Term may be a figure, or a letter, or a combination of
figures and letters, or of letters only, produced by multi
plication or division, or both.
Thus, in the algebraic expression 5 + 2a®6* — a+ gi^; 5, 2a^b*, a,
ah "^y
«2 ,,w
are terms.
An Algebraic Expression is a representation of a number
by any combination of algebraic symbols.
Example. Find the cube root of 27 a — 8 a^ 36+ 36 a^ 12a ^
54a^ + 9a§+27a§ + a66aJ.
The work is conveniently arranged as follows :
EVOLUTION.
91
+
I
1
+
+
•4M
•*
HB«
o
l„
1
<N
C3
o
J
CO
«o
CO
1
1
1
1
**— ^
^
e
C5
05
4
+
r*
k
1
I
1— •
1
CO
1
1
o
1
t^
r^
04
(N
+
+
■an
im
c2
1
<2
1
1
1
5
^
CO
CO
+
+
»«•
HM
1
1
1
1
1
C 8
t I
<M CN
+
Mi
CD
I
I
4
4, ♦ jJ 4, O)
02 CO H cc cw
92 ELEMENTS OF ALGEBRA.
Exercise 34.
Find the cube roots of :
1. x^— Sx^ + 5x^ 3x 1; x^ Saa^+ 5 a^ a^
— S a^ X — a^.
2. 8 0^6 + 48 ax^+60 a^x'^80 a^x^ 90 a^x^+lOSa^x
 27 a^.
3. x^6x^+ 15 x^20 a^+15x^6x+l.
4. 27a^54.a^h + 9a^^+2SaH^3a^b^6ah^h^
5. Sx^+ 12 x^ SOx* 35 x^ + 4:5x^ + 27 x  27,
6. 216 + 3422:2+i7i^4 4.27^6_27^5_i09^108ic.
7. a3  3^25  53+ 8c3+ 6a2c12a&c + 6fe2c
412ac2 125c2+ 3ah\
8. 1  3rK + 62^2  10 rt3 + 12 ic^  12 a;^ + 10 ri;^  6 a;^
4 3 a:8  .^'9.
9. 8 x^  36 x^y + 114 aj*?/^ _ 207 x^y^ + 285 0^23^
225 0^2/5+125 7/6.
10. a^ + 6a^h  Sa^c + 12 a V^  12 abc + 3ac2
+ 8^3 12h^c+ 6bc^c^.
11. x^ + Sx^ySa^y^~lla^f+6xh/^+12xf  8/.
12. 204 rr*2/2  144 a;^ 7/ + 8 / 36 o;^/^ _ 171 ^^s^^s _. 54^6
+ 102 0^2/*.
43. Cube Root of Numerical Numbers. The method for
extracting the cube root of arithmetical numbers is based
upon the algebraic method.
EVOLUTION. 93
Since the cube root of 1000 is 10 ; of 1000000 is 100, etc., it fol
lows that the integral part of the cube root of numbers less than KXX)
has one figure, of nuiubers between 1000 and 1000000 two figures,
and so on. Hence,
I/a point be placed over every third fgure in any number, beginning
with units' place y the number of points will show the number of figures
in the cube root.
Thus, the cube root of 274625 has two figures ; the cube root of
109215352 has three figures.
If the given number contains decimals, the number of decimal
places in the cube root will be one third as many as in the given
number itself. Thus, if 1,11 be the cube root, the number will be
1.367631 ; if .111 be the root, the number will be (3.00i36763i ; if
11.111 be the root, the number will be 1371.706960631. Hence,
The number of points to the left of the decimal point will show the
number of integral places in the root, and the number of points to the
right will show the number of decimal places.
Example 1 . Find the cube root of 778688.
a + 6 = 92
Process. 778688 ( 90 + 2 = 92
The cube of a, or 90, 729000
First remainder, 49688
First trial divisor 3 a", or 3 (90)^ = 24300
3 times the protluctof a and 6, or 3X90X2= 540
Second term b of the root squared, 2*'* =
First complete divisor, 24844
First complete divisor X 6, or 2, 49688
Explanation. There will be two ^gures in the root. Let a f 6
denote the root, a being the value of the number in tens* place, and
b the number in units' place. Then a must be the greatest nmltiple
of 10 whose cube is less than 778688, this is 90.. Subtract a«, or
the cu>)e of 90, from the given number. Dividing the remainder
by 3 a*, or 24.3()0, gives 2 for the value of 6. Add to the trial divisor
3 a 6, or 54(), and 6^, or 4, for the complete divisor. Multiplying by
2 and subtracting, there is no remainder. Hence, 92 is the required
94
ELEMENTS OF ALGEBRA.
root, because we have actually cubed it and subtracted this cube from
the given number and found no remainder.
Example 2. Find the cube root of 897.236011125.
= 24300
1620
36
25956
Process.
Cube of 9,
First remainder,
First trial divisor, 3 times (90) ^
3 times the product of 90 and 6,
6 squared,
First complete divisor,
First complete divisor multiplied by 6,
Second remainder.
Second trial divisor, 3 times (960)2  2764800
3 times the product of 960 and 4, 11520
4 squared, 16
Second complete divisor, 2776336
Second complete divisor multiplied by 4,
Third remainder,
Third trial divisor, 3 times (9640)2 = 278788800
3 times the product of 9640 and 5, 144600
6 squared, 25
Third complete divisor, 278933425
Third complete divisor multiplied by 5,
897.236011125(9.645
729
168236
155736
12500011
11105344
1394667125
1394667125
Let the student formulate a method for arithmetical cube root from
what has been demonstrated.
Note. The notes in Art. 41 are equally applicable to cube root, except that
in Note 1 two ciphers must be annexed to the divisor instead of one.
Exercise 35.
Find the cube roots of:
1. 74088; 34012.224; .244140626.
2. ^mif^; .000152273304.
EVOLUTION. 95
Find to three places of decimals the cube roots of :
3. .64; .08; 8.21; .3; .008; J; ^.
44. Since a* = c^^ = (a*)* = Vo* = VVa,
The fourth root is the square root of the square root.
*
Since a* = a^ '<3 = (aSf = Va* = V 4^,
The sixth root is the cube root of the square root. Hence,
When the root indices are composed of factors, the ope
ration is performed by successive extraction of simpler
roots.
Hote. It is suggested that the teacher use the remainder of this article at
his discretion.
We may find the fifth, seventh, eleventh, or any root of an
expression or arithmetical number if desired, by using the
form for completing the divisor. Thus,
To find the fifth root.
Form, (a + 6)» = a* + (5 a* + 10 a«6 + 10 a*^** j 5 a 6» + 6*) b.
Trial divisor, 5 a*.
Complete divisor, (5 a* h 10 a» 6 + 10 a* &« + 5 a 6» + b*) .
To find the seventh root.
Form, (a+6)' = a7+(7a»+21a»H35a<6«f35a»6»+21a26<+7a6»+6«)6.
Trial divisor, 7 a*.
Complete divisor, {7a*+2la^b+35a*b^+3ba%^2la^b^7ab^+h%
96
ELEMENTS OF ALGEBRA.
Example. Find the fifth root of 36936242722357.
Process.
36936242722357(517
a5 = 5S =
3125
First remainder,
56862427
First trial divisor =:5a'^(a considered
as
5 tens) = 5 (50)* =
31250000
10 a^b (h considered as 1 unit) =
10(50)3 X 1 ::=
1250000
10a2 62= 10 X (50)^X (1)^ =
25000
5 a 63 = 5X (50) X (l)^ ^=
250
6* = (1)^
1
First complete divisor,
32525251
First complete divisor multiplied by
1, 32525251
Second remainder.
2433717622357
Second trial divisor = 5 a^ (a considered
as 51 tens) = 5 X (510)^ = 338260050000
10 a^b{b considered as 7 units)
= 10 X (510)3 X (7) =
lOa^62 = 10 X (510)2 X (7)2 =
5a68 = 5X(510) X (7)3 =
6* = (7)4
Second complete divisor, 347673946051
Second complete divisor multiplied by 7, 2433717622357
9285570000
127449000
874650
2401
Miscellaneous Exercise 36.
Express the nth. roots of:
1. ah^c'^; 52a;3"(^;y + 2«)4"x2%^47/"y"x4"(a;— 2/T
2. Simplify 4: a (Sax y)^ — 5 xhj^ ^2S^ «^ X ak
Find the square roots of:
3. ^x + l^\x^—4x^^l2x^.
4. 2824tt^16al + 9ft" + 4a^
EVOLUTION. 97
5. 162;'^"+16ic74a;8*4x9'' + a:io».
6. a^u~l—4:xhj~^ + 6 — 4 x~^ yi + x~^t/^.
7. 6«cr/;5 + 4 62u;* + a^x^^ ^ \) c^ 12 bcx^  4: abx'.
8. {x'^ + Ax^ + ^ax^ ^ ia^2x^^ ax.
9. a2»+ 2a\6"'+ z'""; « ± 2 ai a;i + x.
Find the cube roots of :
10. 60 x^i/ ^ 4S xf ^27 x^ + lOS x^i/ 90 si^y^ { 8/
80a^^.
11. 24a;*'" 2/2 4 % 3^'^yin_ 6a^'»7/" + a:6m_9g^^n
+ 64/" 56a;3«2^*
12. 15x*  6:ri— 62;6 + 15aj2+ 1 + a:6^20a;8.
13. Su^^2^f + ixi/^f.
14. ^ai6aJJ + 8at^a8 + 27ai + 54a4
+ at + 36 a4 18a2.
Find the sixth roots of:
15. 1215a*1458a6+135a2540a3_l8« + l+729a«
16. .x^+ f6x ?/+ 15jy^ij^Gx^y+ 15 3^1/ 20 a^y^l
17. 160 a3 + 240 a* + 60 a2 + 192 a^ + 64 a«+ 12 a + 1.
18. 2985984; 262144.
Find the eighth roots of :
19. </8+28a3+8a+l + 56a8+70a* + 8aH56«6^28a«.
20. («* + 5*  2 a&8 + 3 a2i^  2 aH)*.
7
98 ELEMENTS OF ALGEBRA. ^
21. Find the 5tli root of 36936242722357.
22. Find the 7th root of 1231171548132409344.
Extract the following roots :
23. y/(a4 + 19 a2 + 25  6 aS _ 30 a),
24. y/[^^ — 2 {m + 71)0^ + (m^ + 4:mn \ n^) x^
— 2mn (m \ n) x + 7/^^ 71^] 2
25. [25 a2 _ 20 a 5 + 4 52 + 9 c2  12 6c + 30 ac]i
26. [27a^ 54a5+ 63 a^Ua^^ 21a2_6a + l]i
27. y/(aj2'" + 2 af"^'^  2 ^'"^^ + ^2» _ 2 aj"+^ + ^).
28. [a6_ 12^5+ 60^4 160^3+ 240 a^192a+ 64:]^.
29. [(«4fe)6'»^34.6^n^(^^j^4m^_^^2rt2V(a+&)2'"ic+8a3V]i
30. y/(^2« ^ 2 a:2«l + 3 3^2—2 _^ 2 a;2'»3 + aj2n4)^
31. y^(8  12a3«i + 6a6"2_a9«3^^
Queries. What signs are given to even and odd roots ? Why ?
What principles govern the signs of roots ? Upon what principle is
the method lor finding the root of a monomial hased 1 How derive
the method for finding the square root of any polynomial '? Why
divide the first term of the remainder hy twice the terms of the root
already found for the next term of the root ? Why add the quo
tient to the trial divisor for the complete divisor ? How derive the
method for finding the cuhe root of any polynomial ? Why divide
the first term of the remainder by three times the square of the root
already found for the next term of the root ? Why add to the trial
divisor three times the product of the terms of the root already found
by the next term, and the square of the next term, for the complete
divisor ?
USE OF ALGEBRAIC S\MBULS. 99
CHAPTER VIII.
USE OF ALGEBRAIC SYMBOLS.
45. Symbols of operation are used to indicate that
algebraic operations are to be performed.
Thus, m f (a  6) indicates that a — 6 is to be added to m ;
m — (a — 6) indicates that a — h is to be subtracted from m. Per
forming the operations, we have,
m\ (a — b) = m + a — b ;
m — (a — h) = m — a + b. Hence,
A plus sign before a symbol of aggregation shows that the enclosed
terms are to be added to what precedes ; as this operation does not
change the signs, the removal of the symbol does not affect the signs.
Removing one preceded by a minus sign changes the sign of each
enclosed term.
Thus, a26[4a66{3ac+(5a263ac + 2 6){]
= a26[4a66{3ac + (5a263a + c2 6)}]
= a2 6[4a66{3ac+(2a46 +c )}]
= a26[4a66{3ac+ 2a46 +c ]
= a26[4a66{6a 46 }]
= a26[4a665a +46 ]
=a26[ 26 o ]
= a26 +26+ a
= 2a
Explanation. Remove the vinculum, subtract and unite like
terras ; then remove the parenthesis and unite like terms ; now
remove the brace, subtract and unite like terms ; finally, removing
the bracket, subtracting and uniting like terms, we have 2 a.
100 ELEMENTS OF ALGEBRA.
Exercise 37.
Simplify :
1. 2al3b+(2b~c)4:c+ {2 a(Shc2h)}'].
2. a — b + c  (a + b — c) — (c — b a).
3. x^ ~[4x^~ {6 x^  (4.r  1)}] (x^ + 4:x^+6x^
H 4^ + 1).
4. ~l0(x\y)lz + x + y3{x + 2y(z+xy)}^
4 4:Z.
5. a[5b{a{5c2cb4b) + 2a{a2bhc)}l
6. 5{a6[a{b c)]} + 60 {6  (c + a)}.
7.2a(3b + 2c)[5b{6cU) + 5c{2a(c+2b)}].
8. 3:r  {?/  [?/  (:r + ?y)  { ;/  (^^ _ ^  ^z)}]}.
9. 3a[2b + ab]^ [Sb 2a + bl
10. {(re 2 ?/ + a; ?/)  (a;  ?/ + ^)}{a: (x2j + xy)}.
11. I a  [ a  {1 «  (2 ft  5 a + 6)}  (f a  3)].
12. f{f(a6)8(&c)}{(60i(c«)}
 I {^  a  («  6)}.
13. 5 {a  2 [ft  2 (ft + a;)]}  4 {«  2[a 2 (a + rr)]}.
14. ft+25{6a~[3& + (8a;2 + &.ya; + 4a)]3/;}
+ 2(1 + «46).
15. 2 (&  f ft) ~ 7 [ft  6 {2  5 (ft  b}}\
16. {f[4(:.i)]}+f{3(a;^)}.
17.  I { L" Q^  h)]} + { I [ (ft  fe)]}.
USE OF ALGEBRAIC SYMBOLS. 101
18. 5 {a  2[6  3 (c H d)]}  4 {a  3 [6  4 (c  </)]}.
19. (al)(a  2)  3a(a + 3) + 2 {(a + 2)(a + 1) 3}.
20. {xz {x'y){y + z)} y[y{x z)\
21. {a^h\c\df^{ahc^ df+ (ab + cd)^
+ (a + bc d)\
22. 7!'{2xy[{x{yz}){x+{yz})\2xy\4.yz}
23. a(a + l)(aH2)(a + 3)6(2aJ)"ta23a + l)2.
24. 571 {(a:— y)a&2;}— 27i{a:(a6)af ?/}  {Zax—{pzr2x)h} n.
25. (a:2 4.2^)^_(a; + y)(a;{7iy}y{^i4).
26. 2aVi — 3 m — [6yi — 6 7i + (2:i2Vi)a] + &X Vy.
27. (9 7?l2 7l2  4 71*) (7m2  7l2)  {3 77? 7?.  2 W^}
{3 771 (/?t2 4 71^) — 2 71 (71* + 3 771 71 — TH?)} 71.
28. 77l2 ( w2 + 7l2)2  2 7^2 ^2 (^ + n) (771  7^)  (tTI^ _ n3)2..
29. i(^^ + iy)(i^Jy)(J^!.y)^?(^f2/^).
30. Y a ^ + 4 ?/') ( J ^  J ?/')  (i :r  3) (J :r + 3)
(4^9) + (f y  3)(f y + 3) (^ .7/2  9).
The use of symbols of aggregation aid in shortening the
work in certain cases in division. Thus,
a 4 (6 + c) ) (6 + c) a^ + (62 I 6 c + c') a   (6 + c) 6 c ( (6 + c) a  6c
(6 + c)a« + (62+ 2 6c + c«) a
+ (  6c )a(6 + c)6
+ (  6c )a(6 + c)6
102 ELEMENTS OF ALGEBRA.
Divide :
31. (6 + c)a2 + (&2 + 32^c + c2)a+(i + c)&c by a + h + c.
32. (a \hf6{a + b) 27 by (a + b) + 3.
33. {x+ijf+ 3 {x + yfz + S{x + y)z^+^ by (x + yf +
2(x + y)z + z^.
34. (x + yf _. 2 (^ + ?/) z + z^ by x ^ y — z.
35. {a + &)3 + 1 by « + J + 1.
46. The converse operation of enclosing any number
of terms of an expression in a symbol of aggregation is
important.
a + m~c\h — n = a + m — c\(h — 71).
a — m — c ] b ~ n = a — (m + c) ^ (b — n).
ax^ — ny + hx^ — cy^= {ax^ — ny) + (hx^ — c y*).
xy — ax'by + ab = (x y — by) — (a x — a b).
Hence, when the signs + and — indicate operation :
(1) Any number of terms may be enclosed in a symbol of aggrega
tion preceded by the sign + , without changing the sign of each term.
(2) Any number of terms may be enclosed in a symbol of aggrega
tion preceded by the sign — , if the sign of each term be changed.
The terms may "be enclosed in various ways. Thus,
am + an — ax — bx + cy — dz=(am — ax) + \an — bx^\\cy — dz\,
or, am\an—ax — bx\cy—dz= (am, + an — ax)—{bx — cy + dz],
or, am + an — ax — bx+cy — dz= {am\an)—{ax\bx)+{cy—dz). Etc.
If a factor is common to each term within a symbol of aggregation,
it may be placed outside as a multiplier. Thus,
ax^ + bx'^~bx^ + dx'^={ax^bx^) + {bx'^+dx^)=x^{ab)+x'^{b + d).
Note. An expression consisting of three or more terms may be raised toa
given power by inspection, by first changing it to the form of a binomial.
Thus, (a + & + c — (^)4 = [(a + J) + (c — d)Y^ = etc.
USE OF ALGEBRAIC SYMBOLS. 103
Exercise 38.
Bracket the last three terms so that each bracket shall
be preceded by a — si<,'n :
I. a:*  a x8  5 a^» + 2 ; m6 + 3 w3 + 3  6 m2.
3. 4:X+'dao(^—{)7^ — bey ^ y; x^—y^—z^\ab\3ac.
4. Express each of the above as binomials, and enclose
the last two terms in an inner brace preceded by a — sign.
Bracket the following in binomials, also in trinomials,
each preceded by a — sign :
5. 2ab — Say\4:bz — 5bx2cd — 3.
6. a— 26 + cz — d — l^z — x—2y+ 2m — n\p—4ahc.
7. 2xSxy{ 4 ^f  5 2:3^2 + ^,p _ ^.y 2
8. a^+3 a*— 4 a^—3 CT^f a —1 ; —2 m—2> 7i+4p— 5 2;— 1—6 y.
9. a n ■\ ab — a c — c X — a X — a y — 3 ab c \ Z xy z.
10. Express the above six examples in trinomials, and
enclose the last two terms in an inner bracket preceded by
a — sign.
II. Expand (tw + 2 71 — j^.
12. Simplify and bracket like powers of x in 2b^—ax
{a:i^\bxnx {a^ ^ 3r.i2} _ {ax'2cx)}.
Queries. Why may a 8ymlx)l of aggregation preceded by a +
sigu be remove<l without changing the signs of the enclosed terms ?
If a symbol of aggregation pi*ecedetl by a — sign be removed, why
change the signs of the enclosed terms?
104 ELEMENTS OF ALGEBRA.
CHAPTEE IX.
SIMPLE EQUATIONS.
47. 3 a: + 5 = 5 a; — 7 is called an Equation. The first
memher or first side is 3^ + 5, and the second member or
second side is 5 x — 7.
x = x, 14 = 14, are called Identities or Identical Equations.
To solve an equation is to find the value of the unknown
number.
The process of solving an equation depends upon the
following axioms:
1. If to equal numhers we add equal numbers, the sums
are equal.
2. If from equal numhers we subtract equal numbers, the
remainders are equal.
3. If equal numbers are multiplied by equal numbers, the
products are equal.
4. If equal numhers are divided by equal numbers^ the
quotients are equal.
Example 1 . Find the value of x in the equation 6 a:— 11 — Sx+lO.
Solution. Subtracting 3 x from each member of the equation
(Axiom 2), we have 6a; — 3a;— ll = 3cc — 3.X+10. Uniting like
terms, 3 a; — 11 = 10. Adding 11 to each member (Axiom 1), and
uniting like terms, we get 3 a; = 21. Dividing both members by 3
(Axiom 4) gives x—1.
Proof. To verify this result, substitute 7 for x in the given equa
tion. Then, 6 X 7  11 = 3 X 7 + 10, or 31 = 31, which is an
identity. Hence, the value of x is 7.
SIMPLE EQUATIONS. 105
Example 2. Solve the equation 2 (a:  8)  3 (9  x) + 5 (a:  1 1)
= 7 3 (a: 17).
Solution. Performing the indicated operations, and uniting like
terms, 10 a; — 98 = 58 — 3 z. Adding 3 x and 98 to each member of
the equation, we have 10 « + 3 a:  98 + 98 = 58 + 98  3 a: + 3 x,
or uniting like terms, 13 x= 156. Dividing both members by 13,
X 12.
Proof. Substitute 12 for x in the given equation.
Then, 2(12«)  3 (9  12^ + 5 (12  11) = 7  3 (12  17),
or, 8 + 9 + 5 = 7 + 15,
or, 22 = 22, an identity.
Therefore, the value of x is 12.
Example 3. Solve the equation 14 — x — 5 (x — 3) (x f 2)
+ (5  x) (4  5 x) = 45 X  76.
Process. Simplify, 64  25 x = 45 x  76.
Subtract 45 x, 64  70 x =  76.
Subtract 64,  70 x =  140.
Divide by — 70, x = 2.
Notet : 1. To verify, that is, to jncne the truth of the result^ substitute the
supposed value of the unknown number in the given equation and thus find if it
satisfies its conditions.
2. In simplifying an equation the student should be careful to notice that
when the sign — precedes a term, in removing the symbol of rggregation, the
sign of each term must be changed.
Exercise 39.
Solve the following equation.s :
1. 6a;+ 1 = 5a:f 10; 11  7z = 18a: 14.
3. 2r+3= ir,(2ar3); 3 (a: 2) + 4 = 4 (3  a:).
4. 7(a: 18) = 3(x 14); 7a:+63a: = 56 + 2.r.
5. 15 (x  1) + 4 (x + 3) = 2{x+ 7).
6. 5  3 (4  a:) + 4 (3  2 x) = 0.
106 ELEMENTS OF ALGEBRA.
48. If we add the same number to each member of an
equation, or subtract it from each member, the results are
equal, each to each. Thus,
Consider the equation x — b = a. Adding b to each side, we get,
X = a { b.
Consider the equation x + b = a. Subtracting 6 from each side,
we have x = a — b.
In each case b is transposed from one side to the other,
but its sign is changed. Hence,
Any term may he transposed from one side of an equation
to the other, provided its sign be changed.
Example. Solve (a: + 1) (x + 2) (x + Q)  (x  2) (x \ 2)
= x^ + 9x^ + 4(7xl){ (2x)(3 + x).
Process. Simphfy, x^ + 8x^ }20x\\6 = x» + 8x^}27x + 2.
Transpose, x^  x^ + 8x^ ~ 8x^ + 20 x27 x = 2  16.
Unite Hke terms, — 7 x = — 14.
Divide by — 7, x = 2.
* Hence, in general,
To Solve a Simple Equation of one Unknown Number. If
necessary, simplify the equation. Transpose all the terms
containing the unknown number to one side, and all other
terms to the other side. Unite like terms, and. divide both
sides by the coefficient of the unknown number.
Exercise 40.
Solve the following equations :
1. 12 X  20 a: + 13  9 2^  259 ; 336 + (3 :c  1 1)
= 2 (5 2:  5) + 8 (97  7 a:).
2. 62; + 4a; = 3:r + 84; 6a: + 2(132^) = 3 (17 a;).
SIMPLE EQUATIONS. 107
3. 2 (a; +2) + 182: = 3(5 + 2;) + 0; 30 a; + 20 2: 15 a;
+ 12 a; = 2820.
4 9(2; _ 1) + 2 (a: 2) =10 (2 a:); 2 (a: + 2) (a:  4)
= a;(2a;+ 1)2L
5. 6y2(94?/) + 3(52/7) = 10 2/(4 + 16y+35),
and verify.
6. 2y(4yl) = 5y(//+l); 56+ 21 a; 8 (2 a; 1)
= 62.
7. 10 [224 {xV 192)] = 7 (28 + 3a:); 9 (7 + ^y)
4[9(2.y)] = 252y.
8 25 a: V.)  [3{4a:3}] = a: — (a: — 5), and verify.
9 20(2~a;) + 3(a:7)2[a;+93{94(2a:)}] = 1.
10. (y2)(77/) + (7/5)(y+3)2(y^l) + 12 = 0.
11. 4 (v/ + 5)2  (2 y + 1)2 = 3 (y  5) + 180 ; 2.25 x
 1.25 = 3 a: + 3.75.
12. .15?/ + 1.575 .875?/ = .0625 2^.
Query. In transposing, why change the signs ?
49. Known Numbers are represented by the first letters
of the alphabet, and by figures ; as, a, b, 2 c, 6.
Unknown Numbers are usually represented by the last
letters of tlie alpliabet; as, x, y, z.
An Equation i.s a statement that two expressions repre
sent the same number.
An Identical Equation, or an Identity, i.s one which is
true for all values of the letters which enter into it ; as,
{a + z) (a — a:) = a2 — 3^.
108 ELEMENTS OF ALGEBRA.
The Roots of an equation are the values of the unknown
numbers.
The Degree of an equation is the power of the unknown
number, and is determined by the greatest number of un
known factors in any term.
Thus, X — y = 6 is an equation of the Jirst degree ; 4x^ + 5y = 3
and 5 xy + 2 — 3 X ave equations of the second degree.
A Simple Equation is an equation of the first degree.
Miscellaneous Exercise 41.
Solve the following equations :
1. 5(7 + 3:?/)(23/3)(l2?/)(2,y3)2 + (5 + 2/)  0.
2. {22j+lf + {27/lf=UyQf4:)+57, and verify.
3. 1.5(26i/51)12{l3y)=7Sy2[5y2.6{l.Sy)].
4. .6a:.7:r + .752:.875:r + 15 = 0; .6y{.lSyM)
= .2 2/ + 4.45.
5. 30 ^  3 [30 z{2z 5)] = 5{2z 57)  50.
6. 10 (^ + 10)  18 (3 2  4) + 5 (3^2) (2 z  3)
= 30 2;^ — 16, and verify.
7. 4.Sy2{.72yM) = 1.6?/ + 8.9; .5x.3x.25
= .25a:l.
8. .2 7/ .16?/ = .6 .3; .5y  .2y = .3y 15.
9. 5.6 y f .25 y+ .3y = y'S; .6 y \ .25  A y = l.S
.75?/ .3.
10. 3)x .25 (x2)~ .3 (3x{ 12)? = 41.
PROBLEMS LEADING TO SIMPLE EQUATIONS. 109
CHAPTER X.
PROBLEMS LEADING TO SIMPLE EQUATIONS.
50. The beginner will find the model solutions of great benefit
ill forming statements, and he should give them careful consideration
before attempting to solve any of the problems in each set.
Exercise 42.
1. A father is 35 and his sod 8 years old. In how
many years will the father be just twice as old as the
son ?
Solution. Let x — the number of years required.
Then x + 35 = the number of years in the father's age x years
from now,
and X f 8 = the number of years in the son's age x years
from now.
By the conditions of the problem, at the expiration of x years
twice the son's age, or 2 (x + 8), equals the father's age, or a: f 35.
Hence, the equation 2 (x + 8) = x + 35, or 2 x + 16 = x I 35.
Transposing and uniting like terfns, x = 19.
2. One number exceeds another by 5, and their sum is
29. Find the numbers.
3. The difference of two numbers is 14, and their sum
is 48. Find the numbers.
4. A father gave S200 to his five sons, which they are
to divide according to their ages, so that each elder son
shall receive $10 more than his next younger brother.
Find the share of each.
110 ELEMENTS OF ALGEBRA.
5. A father is four times as old as his son ; in 24 years
he will only be twice as old. Find their ages.
6. Divide 50 into two parts, so that three times the
greater may exceed 100 by as much as 8 times the less
falls short of 120.
Solution. Let x = the greater part.
Then 50 — x = the less part,
and 3 X = three times the greater part ;
also, 8 (50  x) = eight times the less part.
But, 3 X — 100 = the excess of three times the greater part over
100;
also, 120—8 (50 — x) = the number that eight times the less lacks of
120.
By the conditions, 3 a:  100 = 120  8 (50  x).
Therefore, . x = 36, for the greater part,
and 50 — a; = 14, for the less part.
7. Twentythree times a certain number is as much
above 14 as 16 is above seven times the number. Find
the number.
8. A is five years older than B. In 15 years the sum
of their ages will be three times the present age of A.
Find the age of each.
9. A is 25 years older than B, and A's age is as much
above 20 as B's is below 85. Find their ages.
10. The sum of the ages of A and B is 30 years, and
five years hence A will be three times as old as B. Find
their ages.
11. The difference between the squares of two consecu
tive numbers is 121. Find the numbers.
PROBLExMS LEADUSG TO SIMPLE EQUATIONS. Ill
Solution. Let x = the less number.
Then will jc f 1 = the greater number,
x^ = the stjuare of the less number,
and (x I 1)*^ = the stpuire of the greater number.
Then (x + 1)^  x'^ = the ditlerence of the scjuaie numbers.
But 121 = the difference of the squares.
Hence, (x + l)^  x = 121.
Therefore, x = 60, the less number,
X h 1 = 61, the greater number.
12. Find three consecutive numbers whose sum is 27.
13. The difference of two numbers is 3, and the differ
ence of their squares is 21. Find the numbers.
14. Find a number such that if 5, 15, and 35 be added
to it, the product of the first and third results may be equal
to the square of the second.
15. I sold a cow for S35 and half as much as I gave for
it, and gained SIO. Find the cost of the cow.
16. A had four times as much money as B ; but, after,
giving B $16, he had only two times as much as B. How
much had each at first ?
Solution. Let x = the number of dollars that B had at first.
Then 4x = the number of dollars that A had at first.
But 4x — 16 = the number of dollars that A had after giving
B^16,
and X + 16 = the number of dollars B had after receiving
^16 from A.
By the conditions, 4 X 16 = 2 (x + 16).
Therefore, x = 24, the number of dollars that B had,
and 4x = 96, the number of dollars that A had.
17. A father is 3 times as old as his son ; four years
ago the father was 4 times as old as bis son then was.
Find their ages.
112 elp:ments of algebra.
18. One number is two times another; but if 50 be
subtracted from each, one will be three times the other.
Find the numbers.
19. A has $26.20 and B has $35.80. B gave A a cer
tain sum; then A had four times as much as B. How
much did A receive from B ?
20. If 288 be added to a certain number, the result will
be equal to three times the excess of the number over 12.
Find the number.
21. A farmer has grain worth $0.60 per bushel, and
other grain worth $1.10 per bushel. How many bushels
of each kind must be taken to make a mixture of 40
bushels worth $0.90 a bushel?
Solution.
Let X = the number of bushels required of the f 0.60 grain.
Then 40 — a: = the number of bushels required of the $1.10 grain;
and Y®^*\j X — the number of dollars in the cost of the $0.60 grain;
also, 1.10(40a:) — the number of dollars in the cost of the $1.10grain.
Hence, ^^^ a; f 1.10 (40 — x) — the number of dollars in the total
cost of the mixture.
But the cost of the mixture is to be $36. Hence,
^%x\ 1.10 (40 a:) = 36.
Therefore, x = 16, the number of bushels of the $0.60 kind,
and 40  X = 24, the number of bushels of the $1.10 kind.
22. A merchant has two kinds of vinegar : one worth
$0.35 a quart and the other $1.25 a gallon. From these
he made a mixture of 63 gallons, worth $1.30 a gallon.
How many gallons did he take of each kind ?
23. A merchant has a mixture of 88 pounds of 13 and
11 cent sugar, which he sells at 12 cents per pound.
How many pounds of each kind are there ?
PROBLEMS LEADING TO SIMPLE EQUATIONS. 113
24. I bought 24 pounds of tea of two different kinds,
and paid for the whole $9. The better kind cost $0.65
per pound, and the poorer kind $0.35 per pound. How
many pounds were there of each kind ?
25. A grocer having 75 pounds of tea worth $0.90 a
pound, mixed with it so much tea at $0.50 a pound that
the combined mixture was worth $0.80 a pound. How
much did he add ?
Remarks. No general method can be given for the solution of
problems. ^
The beginner will find that his principal difficulty in solving a
problem consists in forming the equation of conditions, and in order
to overcome this, much will depend upon his skill and ingenuity.
The statement of a problem consists in translating its conditions
into algebraic symbols and ordinary language. Many times the be
ginner fails to form a correct statement, because he does not under
stand what is meant by the ordinary language of the problem. If he
cannot assign a consistent meaning to the words, it will be impossi
ble for him to express their meaning in algebraic symbols. It often
happens that the words appear to be susceptible of more than one
meaning. In such caj^es the student should express the meaning that
seems most reasonable in algebraic symbols, and obtain the result to
which it will lead. Should such result be inadmissible, the student
should Uy another meaning of the words.
The student must depend upon hus own powers^ and should he at
times be perplexed, he must not be discouraged, since nothing but
patience and practice can overcome the difficulties and give him
readiness and certainty in solving problems. He must study the
iT»eaning of the language of the problem, to ascertain the unknown
numbers in it. There may be several such numl^rs, but oftentimes a
little skilful manipulation will enable one to express all of the un
known numbers in terms of some one of them. Select the one by
which this can be most easily done and represent it by some one of
the final letters of the alphabet.
Among the following problems no doubt the beginner will find
8
114 ELEMENTS OF ALGEBRA.
some which he can readily solve by arithmetic, or by guessing and
trial; he may thus be led to undervalue the power of algebra, and to
regard its aid as unnecessary. In reply, as the student advances lie
will find that by the aid of algebi'a he can solve not only all of these
problems, without any uncertainty or guessing, but those which
would be exceedingly difficult, if not altogether impossible, if he
depended upon arithmetical processes alone.
26. A's age is six times B's, and fifteen years hence A
will be three times as old as B. Find their ages.
27. A is three times as old as B, and 12 years since he
was fiv^ times as old. Find B's age.
28. A father has three sons; his age is 60, and the
joint ages of the sons is 46. How long will it be before
the joint ages of the sons will be equal to that of the
fathei' ?
29. If yon walk 10 miles, then travel a certain distance
by train, and then twice as far by coach, and the whole
journey is 70 miles, how far will you travel by coach?
30. A is twice as old as B, and seven years ago their
united ages amounted to as many years as now represent
the age of A. Find their ages.
31. After 136 quarts had been drawn out of one of two
equal casks, and 80 gallons out of the other, there remained
just three times as much in one cask as in the other.
Find the contents of each cask.
32. Find the number whose double increased by 1.2
exceeds 3.65 by as much as the number itself is less than
8.65.
33. Find three consecutive numbers such that if they
be diminished by 10, 17, and 26, respectively, their sum
will be 10.
PROBLEMS LEADING TO SIMPLE EQUATIONS. 115
34. Two consecutive numbers are such that one fourth
of the less exceeds one tifth of the greater by 1. Find the
numbers.
35. There are two consecutive numbers such that one
fifth of the greater exceeds one tenth of the less by 3.
Find tliem.
36. Find a number such that the sum of its half and
its fourth shall exceed the sum of its fifth and its tenth by
45.
37. Find a number such that the sum of its half and
its fifth shall exceed the difiference of its fourth and its
tenth by 110.
38. If a watch and chain are worth $185, and the
watch lacks $19 of being worth two times the cost of the
chain, find the cost of each.
39. If silk costs 6 times as much as linen, and I buy
22 yards of silk and 28 yards of linen at a cost of $52,
find the cost of each per yard.
40. A man gave 17 boys $3.31, giving to some 13 cents
each and to the rest 23 cents each. How many received
23 cents ?
41. I paid a bill of $1.53 with 39 pieces of money,
some 3cent and the rest 5cent pieces. How many of
each did it take ?
42. A son earns 37 cents per day less than his father,
and in 8 days the father earns $6.08 more than the son
earns in 5 days. Find the daily wages of each.
43. How many 10cent pieces and how many 25cent
pieces must be taken so that 95 pieces shall make $12.35?
116 ELEMENTS OF ALGEBRA.
44. Divide $112 into two parts, so that the number of
fivecent pieces in one may equal the number of threecent
pieces in the other.
45. A sum of money consists of dollars, twentyfive cent
pieces, and dimes, and amounts to $29.50. The number
of coins is 55. There are twice as many dimes as quarters.
How many are there of each kind ?
46. A sum of £8 17 s. is made up of 124 coins, consist
ing of florins and shillings. How many are there of each?
47. A bill of £4 5s. was paid in crowns, halfcrowns,
and shillings. The number of halfcrowns used was four
times the number of crowns and twice the number of shil
lings. How many were there of each ?
48. A bill of £48 J was paid with guineas and half
crowns, and 12 more halfcrowns than guineas were used.
How many were there of each ?
49. A company of 84 persons consists of men, women,
and children. There are three times as many men as
women, and five times as many women as children. How
many are tliere of each ?
50. The sum of three numbers is 263. The first is 3
times the second, and the third is 23 more than 5 times
the sum of the other two. Find the numbers.
51. A farmer wishes to mix 660 bushels of feed, con
taining oats, corn, rye, and barley, so that the mixture
may contain two times as much corn as oats, three times
as much rye as corn, and four times as much barley as
rye. How many bushels of each should be used ?
PROBLEMS LEADING TO SIMPLE EQUATIONS. 117
52. Divide $2590 into two sucli parts that the first at
7% simple interest for 8 years may amount to the same
sum as the second in 5 years at 8 %.
Note. The character % is sometimes used for the term "/>er cent." Per
cent is used by ellipsis for rate per cent. Thus, au allowauce of 7 on a hundred
is at a rate of .07, aud the rate per cent is 7.
53. $330 is invested in two parts, on one of which
15% is gained, and on the other 8 % is lost. The total
amount returned from the investment is S345. Find the
investment.
54. A man has $ 7585. He built a house, aud put tho
rest out at simple interest for 18 months; 40% of it at
5 % and the remainder at 6 %. The income from both in
vestments is $211.26. Find the cost of the house.
55. In a certain weight of gunpowder the saltpetre was
4 pounds less than half the weight, the sulphur 5 pounds
more than a fifth, and the charcoal 3 pounds more than
a tenth. Find the number of pounds of each.
56. A company of 266 persons consists of men, women,
aud children. Tlie men are 14 more in number than the
women ; the children 34 more than the men and women
together. How many are there of each ?
57. I bought 16 yards of cloth, and if I had bought one
yard less for the same money, each yard would have cost
$0.25 more. Find the cost per yard of the cloth.
68. A and B, 85 miles apart, set out at the same time
to meet each other; A travels 5 miles an hour aud B 4
miles an hour. How far will each have travelled when
they meet ?
118 ELEMENTS OF ALGEBRA.
59. $330 is loaned for nine months in two parts ; on
one 15 % per annum is gained, and on the other 8 %
per annum is lost. The total amount from the loan is
$364.25. Find the amount in each loan.
60. A boy has a certain sum of money, lie borrowed as
much more, and spent 12 cents; he again borrowed as
much as he had left, and spent 12 cents ; again he bor
rowed* as much as he had left, and spent 12 cents ; after
which he had nothing left. How much money had he at
first?
61. A carriage, horse, and harness are worth $720. The
carriage is worth eight tenths of the value of the horse, and
the harness six tenths of the difference between the value
of the horse and carriage. Find the value of each.
62. A boy sold half an apple more than half his apples.
Again he sold half an apple more than half his remaining
apples. A third time he repeated the process; and he had
sold all his apples. How many apples had he ?
Algebra is the science which treats of algebraic liumbers
and the symbols of relation.
Algebra, like arithmetic, is a science which treats of numbers. In
arithmetic the numbers are positive and represented by figures. In
algebra the letters of the alphabet or figures are used to represent
numbers, and they may be positive or negative, real or imaginary.
Algebra enables us to prove general theorems respecting numbers,
and also to express those theorems briefly.
FACTORING. 119
CHAPTER XL
FACTORING.
51. A Factor is one of the makers of a number.
Thus, since 5 with the aid of 4 and by the process of multiplica
tion makes 20, 5 is a factor of 20.
A factor is also a divisor, but it is considered a divisor when it
separates a number into parts, not when it helps to make up a
number.
Note. Unity cannot be a factor.
Factoring is the process of separating an expression into
its factors.
Example. Find the factors of 12 a« 6 x^.
Solution. The prime factors of 12 are 2, 2, and 3. The factors
of a' are a, a, and a. The factors of x^ are x and a:*.
Therefore, l2a*bxi = 2X2X3xaXaXaXbXxXxK
Hence, a» a direct result of the principle that monomials are mul
tiplied by writing the several letters in connection, and giving each
an exponent equal to the sum of the exponents of that letter in the
factors.
To Factor a MonomiaL Separate the letters into any number
of factors, so that the sum of all the exponents of each factor shall
make the exponent of that factor in the given expression ; also sepa
mte the numerical coefficient into its prime factors.
Exercise 43.
Separate into factors with integral exponents :
1. Ua^l^x; Wo^t/^; 15ah^(^; 20ab(^; Soa^f:!^^;
120 ELEMENTS OF ALGEBRA.
Separate into two equal factors :
2. UaH'^; da^f; SI a^ b^ x^"" y^"" ; 169a"5.
Eemove the factor 2 a^ bi from :
3. Sa^b; Qabx; IQab^c^; 10 a'H^ x^y^.
Separate into three factors, also into four :
4. cc ; m^ " ; a" ; xi ; x^.
52. Example 1. Factor a^x  Sa^x^.
Solution. Dividing the expression by a^ x, we have a — 3 aj.
Hence, a^ x — 3a^x = a'^x {a — 3 x).
Example 2. Factor 5 a'^b^x^  15 ab^x^ + 20 b« x^.
Solution. By examining the terms of the expression we find
that 5 b^ x^ is a factor of every term. Dividing by this common fac
tor the other is fomid. Hence, the factors are 5 h^x^ and a^x — Sax
+ 4 6.
.. 5aH^x^l5ab^x»\20b^x^ = 5b^x^(a^x3ax + 4b).
Hence,
When the Terms of a Polynomial have a Monomial Factor.
Divide each term of the expression by the common factor. The
divisor and quotient will be the required factors.
Exercise 44.
Factor the following :
1. 7n^ + n; 4: a^b + ab^c+ Sab; Sa^ 12 a^.
2. ax — bx+cx;S9a^7/{57x^y^.
3. 5x^ + Sa^x^; 72 h^ x'^y^  84:b^ x y^  9Q a b a^y^.
4. 924 «2 x^y^'z 1 178 a x"" y z"" + 1232 a^ x"" y'^ z\
FACTORING. 121
5. 4:aH(J0aI^+20abc{SaH*x^+UahyZ6aHcx^,
6. 2 xi y ^ a b X y + c a^ y^ ; 5 x^ + 10 x^ — 15 xi.
53. In certain Trinomiala, of the form x^ \ ax ^ b, where a
and b represent any numbers, either integral, fractional, positive, or
negative, it is possible to reverse the operation of Art. 25, and sepa
rate the expression into the product of two binomial factors. Evi
dently the first term of each factor will be the square root of a;*, or x;
and to obtain the second terms of the factors, /ind two numbers whose
algebraic product is the last term, or b, and whose sum is the coefficient
ofx, or a.
Example 1. Factor x^ + 21 a: { 110.
Solution. Evidently the first term of each factor will be x. The
second term of the factors must be two numlnirs whose product is
110 (the third term), and whose sum in 21 (the coefficient of a;). The
only two numbers whose product is 110 and whose sum is 21 are 10
and 11. Therefore, z« + 21 1+ 1 10 = (a: + 10) (x + U).
Example 2. Factor x^ + x 132.
Solution. Evidently the first term of each binomial factor will
be X. The second term of the two binomiul factors must be two
numbers whose algebraic product is — 132 and whose sum is f 1
(the coefficient of x). The only two numbers whose product is — 132
and whose sum is f 1 are 4 12 and — 11. Therefore, x^ \ x — 132
= (X \ U) (x  U).
Example 3. Factor y^  5cy — 50 c^
Solution. Evidently the first terra of each binomial factor will
be y. The second term of the two binomial factors must Ijc two
numbers whose product is — 50 c* and whose sum is —5c (the coef
ficient of y). The only two numbers whose pro<luct is — 50 c* and
whose sum is —5c are + 5 c and — 10 c. .. y^ — 5 cy — 50 c*
= (yl5c)(y10c).
122 ELEMENTS OF ALGEBRA.
Example 4. Factor x^y^ — (jn — n) xij — m n.
Solution. Evidently the first term of each binomial factor will
ho, xy.. The second term of the two binomial factors must be two
such numbers whose product is —nin and whose sum is — (in — n).
The only two numbers whose product is — m n and whose sum is
— (m — n) are + n and — m.
.*. x^y'^ — (in — n) xy — m n = (x y \ n) {xy — m). Hence,
I. If the Coefficient of the Highest Power is Unity. For the
first term of each factor take the square root of one term of the trino
mial ; and for the second term of the factors, such numbers that
their algebraic product will be another term of the trinomial, and
their sum multiplied by the first term of either factor will be the
remaining term of the trinomial.
Exercise 45.
Factor the following :
1. ^24.19^+88; .^272^+ 12; ft8_20a4 + 96.
2. 2:2 + 35 a; + 216; 52(^2 245c + 143.
3. ft* ¥ + 37 a2 62 + 300 ; a^ + 5 ah  66 h\
4. a?ly^oah24. ft* + 15*2 + 44. a^ + 17 ft^ + 60.
5. a^ydahclQc^; a^2a'^120; yi2+.8 ^i + l.o.
7. ^2 _ 15 a; + 44; ^„2 + .i_i. ,„, ^ i^_ . ^2 _ 11 ^^ _ 26.
8. 130 + 31 ft & + ft2 /;2 . ,,2  20 ah x+ 75 h'^ oj^ ; y^
+ 6 a;2 7/2 _ 27 2)4; 1 + 13 ^ + 42 :r2 ; ^;i2  15 a m + 06 a^.
9. ft2  lSax7j  243 x^ y^; (x + yf + 5 {x +. ?/) + 4.
10. 40 ft2 2,2 _ 13 ,^ 7, + 1 . (^ _ 5)2 + (^ h)2.
FACTORING. 123
11. (xyfd{xy)l0;ay^+54x]'729', 20429^^+ .,4.
12. (a hiy ^9 (a + i)2 + 8 ; 2;*"  (6 +'m) x"^"" ^ b vi.
13. a2  10 a V^v  39 h^c^ x^ ^ {a  h) x""  a b.
14. a^9xij70f', a;2..c; 2;*''43ic2n^46Q
15. 2^{iax^^a^; x*a^x'^4e2a\
16. x^i/^Sx7/154:; a^" x*"' + 14 a" x^"' 2/" + 33 y^.
17. a.>2 ?/2 _ 28 a" i" a;?/ + 187 a^ b^  x^^\x + '^^V
18. a;*"//" + 20a'"^/'";;t2''7/2n ^ 5irt2«.j2m. (^ ^ ^^^em
 7 a*" (a; + ?/)3  98 a^" ; n* + .01 n^  .011.
19. ^+^x^\; u;2+2^2/~.21y2; «4+_8^^2+ j^.
By an extension of the foregoing principles we may factor some
trinoniials, of the form c^x^ + ax \ hd^ where the coefficient of a:^ is
a perfect square. Thus,
Example 20. Factor 4 a:« + 4 a: ^ 3.
Solution. The first term of each binomial factor will be the
square root of 4 x^. The second term of the two binomial factors
must >>e two numbers whose product is — .3 and whose sum multiplied
by 2 a; is + 4 x. The only two numbers whose product is — 3 and
whose sum multiplied by 2 a: is + 4 a; are + 3 and — 1 .
.. 4a:« + 4a;3=(2x + .3)(2a:l).
21. 4a.^10:r + 6; 9a^»27a;+18; 4a'2+16aa;+12a2
22. 9 rt2 + 30 « 5 + 24 J2; 16 a^«  20 a a; + 6 ««
23. 25.xiO'"2r5"'rt''Ja2; 36(at)* + 12(a6)*" + *
 143 (a  h)\
124 ELEMENTS OP ALGEBRA.
54. We may factor some trinomials of the form ax^ + bx ■{ c.
Thus,
Example 1. Factor 8 a;^  38 a;+ 35.
Solution. The first term, 8x% is the product of the first terms of
the binomial factors. The last term, 35, is the product of the second
term of the two binomial factors. It is evident that the first term of
each binomial factor might be ±2 a: and ±.4x, or ±Sx and ±a:;
also the last terms of the two factors might be ± 7 and ± 5, or db 35
and ± 1. From these we must select those that will produce the
middle term, —38 x, of the trinomial. Since (+2x) X (— 5) + (+ 4 cc)
X (— V) = — 38 X, we must take + 2 a: and + 4 x for the first terms,
and — 7 and — 5 for the corresponding second terms of the two bino
mial factors. Therefore, 8 a;^  38 a; + 35 = (2 a;  7) (4 ic — 5).
Example 2. Factor 6x* ~ bx'^y^ — 6 y\
Solution. Take + 3 a;^ and + 2 x^ for the factors of 6 x*, and
+ 2 y'^ and — 3 2/^ as those of — 6 ?/*. We now arrange them in bino
mial factors, so that the algebraic sum of their cross products shall be
 5 a:2 1/2. Since (+ 3 a^) X ( 3 7/2) + (+ 2 x^) X (+2y^) = 5x^ y\
+ 3 a;2 and + 2 a;^ are the first terms, and + 2 y'^ and —3y^ are the
corresponding second terms of the factors. .•. 6x^ — 6 x^y^ — 6y*
= (3 a:2 + 2 ?/) (2 a:2  3 y^). Hence,
II. If the Coefficient of the Highest Power is not Unity
Arrange the trinomial in descending powers of a common letter.
Select factors of the extreme terms and arrange them in binomial
factors, so that the algebraic sum of their cross products shall be the
second term of the trinomial.
Exercise 46.
Factor the following :
1. 4 x^ + l.S a: + 3; 4 2/2 _ 4 y _ 3; 12 a^ \ a^a^  x\
2. S + llx4:X^; Sx^'22xy21f; ^o?'x^ + ax^l.
FACTORING. 125
3. 8 m6  19 m3 _ 27; 15 a^ _ 58 a + 11 ; 6 a2 ^. ^ ah
 3 62; 2 //<2 _ 13 m 71 + 6 n2 ; 3 a^» + 7 a; + 4.
4 24 + 37a72a2; 15^:24. 224a: 15; 45a;6a:2.
5. 6a^» 19.ry + 10/; % a^ ^^ U xy  Ibf) lb a^
77 a; +10; 24 a,^ + 22 2; 21 ; lla2 + 34a4 3
6. 1833a;+5ar'; Ga:2_7a,2^_3^. 5 + 32;« 21a;2.
7. 24a,'229a;y4y2; ea^^+lQ^rayn^y^^m
8. 2(a;+y)2+5(a;+y)(m + ?i)42(m + n)a; 2«2+a;28.
9. 2(x+yfl{x^y){a^h) + Z{a + hf l^x'j^^x^.
10. n{xyf''l'6x'^y^{xijf''^2x^'^f) 27a2+6al.
11. 8 a2" f 34 rt» (2:  yY^ + 21 (a:  ;/)2'»».
55. A trinomial is a perfect square when two of its terms are
positive, and the third term is twice the product of their .square roots.
Such trinomials are particular forms of I., and their binomial factors
are equal.
Example. Factor 4 z^ + 44 a:y + 121 y\
Solution. The first term of each binomial factor will be the
8C[uare root of 4 x^, or 2 a: and 2x, For the second terms of the bino
mial factors tjike the square root of 121 ?/*, or 11 ?/ and 11 y. Since
the terms of the trinomial are positive the factors are 2 x + 1 1 y and
2x + lly. Therefore,
4x« + 44xy4 121 y2= (2x+lly)(2x+ 11 y)
= (2x4lly)^. Hence,
III. If the Trinomial is a Perfect Square. Arrange the tri
nomial .according to the powers of one letter. For one of the equal
factors, fin<l the square roots of the first and last terms, and connect
these roots by the sign of the second term.
126 ELEMENTS OF ALGEBRA.
Exercise 47.
Factor the following :
+ 225&2c2; a64a4 + 4a2
2. 49 m6 140 m%2 +100714; si x"^ 9/ 126 a^xhj + 49 a^.
3. 7n}^—2m^nin'^; l10mn + 25m^n^; x'^+2x^ix^.
4. {a + hf + 16 (a + ^>) + 64 ; 7?i2 + 18 m + 81.
5. 4a*a^20a2^7/ + 25a;42/^ o61aH^c^16ahcdmn
+ 4:d^m^'n?; 121 7/2,27^4  220 mn'^2^ + 100^2
6. 225 X^  30 a;2 7/2 + ^4 . 4 ^4« _ 4 ^2n ^m _^ ^2m^
7. 49 m27i2 + 2^ m n^ + ^71*; ^j2 + ^ + 1.
8. ^^a^ + ^^^W + \a^h^ a^c + 6a^h^e+9Wc.
9. 9^:2 _ 3:^7/ + ^7/2; (m  7O2 + 2 (7?z,  ^i) + 1.
10. {ci?af\.6{a^a)+ 9; 4 (a; + 7/)2 + Jg + a? + 7/.
11. at + 6l — 2 aHf ; 7?i — 2 ??ii + 1 ; ??z2 71 + 77^ 7i2 — 2 77it 77!.
12. x\2 x^y^ + 7/ ; 7?i2 n + a2— 2 a m n^ ; 4 ic+ 1 2 71 a:^ + 9 7i2.
13. (a + 2>)2"10(a + &)"c + 25c2;  ^5m_^ _i_6___ 11 J.«.
56. Example. Factor 8 a;^  27 i/^.
Solution. Evidently (Art. 34) 2 re  3 y is a divisor of 8 x^  27 1/^.
Dividing 8a:^ l)y2x, we have 4x^, the first term of the quotient.
Divide 4 rc^ by 2 oj, multiply the result by 2y, and we have Qxy, the
second term in the quotient. In like manner we find 9 .y^ for the last
term in the quotient. Hence, the quotient is ^x^ + Qxy + ^ y^.
Therefore, the factors of the binomial are 2ic — 3?/ and 4x'^+ 6a:?/ + 9?/2.
FACTORING. 127
Since the dividend is equal to the prcKluct of the divisor and quotient,
:i* — 27 y* = (2 X — 3 y) {4 x^ h 6 xy + d y^). Hence, in general,
When a Binomial is the Difference of Two Equal Odd
Powers of Two Numbers. Cunsider the binomiid a dividend,
and find a divisor and quotient by inspection (Art. 34) The divisor
and quotient will be the required factors.
Exercise 48.
Factor the following :
1. ld'i'Sa^; 82^7297/) 216x^a^
2. 2^,/  aH^; x^ 1; 243 a^  b^; a^h^  m^
3. 216 d? — 343 ; 3 a; — 81 a:*. Suggestion. Remove the
monomial factor 3x first.
4 a}^  1024 6^0 ; 729 x^  1728 f\x^ y\
5. 135a;*320ar2; 2an64a&; a;^7/f.
6. a6552:6/; 64^6125^3; a:3n_^«
57. Example. Factor 729 + a«.
Solution. Since 729 is the 6th power of 3, 3* + r2 (Art. 36) is
a divisor of 729 + a' Dividing 729 by 3^* we have 3*, the fiist term
in the quotient. Divide 3* by 3', multiply the result by a'*, and we
have 3* a*, the ."^econd term in the quotient. In like manner we
find a* for the last term in the quotient. Honcc, the quotient is
3*  3"^a2 + n*. Therefore, the factors of the binomial are 3* f a*
and 3* — 3^ a* + a*. Since the dividend is efpial to the ]>roduct
of the divisor and quotient, 729 + (i« = (9 + a^) (81  9 a* + a<).
Hence, in general,
When a Binomial is the Sum of Two Equal Odd Powers of
Two Numbers. Let the student supply the method (See Art. 36).
128. ELEMENTS OF ALGEBRA.
Exercise 49.
Factor the following :
1. d2a^ + 1; 1 + x^; a^ + y^; x^^ + i/^.
2. a^ + 128; x^ + 729 ^/^ 64 2^6 + y^
3. aH^ + 2^10^10; x^ +64/; 1000 a;3 + 1331 ^/^
4. a^l8 + yS J 135^5 _!_ 320 :i;2. ^24 _. ^24,
5. 2;5 + 7/5; j;15 + ^6. ^5^5 +^5^5. ^21 ^ J54,
6. a54 + 654. 1 4. ^12. ^^n __ ^6m. ^f _^ ^ f ^
7. ai2» + &9'" . 32 rj> h^ c^ + 243 a^ ; 1024 a^ + h^^.
8. 64 a;6 + 729 a^ ; yig a^ + g\ je ; (^2 _ J c)^ + 8 h^c^.
58. Example 1. Factor 25 a;^  64 2/*.
Solution. The square root of the first term is 5 x^, and of the
last term 8 y. Hence, since the difference of the squares of two num
bers is equal to the product of the sum and difference of the numbers
(Art. 26), 25 a;2  64 3/2 = (5 x + 8 ?/) (5 a;  8 y).
Example 2. Factor {6 a  4)^  (3a + 4x  4)2.
Solution. The square root of each term of the binomial is 5 a — 4
and 3 a + 4 ic — 4. Adding the results for the first factor, we have
8 a 4 4 X — 8, or 4 (2 a + a; — 2). Subtracting the second result from
the first for the second factor, we have 2a — 4x, ov 2 (a — 2 a:).
Hence, the factors are 4 (2 a + a: — 2) and 2 (a — 2x).
Process.
(5a4)2(3af4a;4)2 = [(5 a4)+(3 a4 .'c4)] [(5 a4)(3 a+4 x4)]
= [5a4 + 3a + 4a;4][5a43a4a; + 4]
= [8a + 4a;8][2a4a:]
= 4[2a + a:2][2(a2a:)]
= 8(2a + a:2) (a2x).
FACTORING. 129
A binomial expressing the diiference between two e<iUttl even
powere of two numbers may often be separated into several factors.
Thus,
Example 3. Factor a" — i/".
Solution. The square root of each term of the binomial is 3* and
y*. Adding these results for the first factor, we have x^ } y^. Sub
tracting the second result from the first for the second factor, we have
x^ — y*. Similarly the factors of x* — y* are ar* + y* and x* — y^. In
the same way the factors of z* — //* aie x^ + y^ and x^ — y"^. Finally
the factors of x^ — j/* are x + y and x — y. Hence, the factors of the
binomial are x* + y*, a:* I y*, x* + y\ x + y, and x — y.
Process, x^' — y^^ = (x^ + y^) (x^ — y^)
= (x^+y^)(x^ + y*)(x*y*)
= (x^ + y^) (a:* + /) (x2 + y2) (x2 1/2)
= (a:«+/) {x*iy*) (xH/) (x + y) (xy).
Hence, in general,
When a Binomial is the Difference of Two Equal Even
Powers of Two "Numbers. Find tlie square root of eacli term of
the biuouiial ; add the results for one factor, and subtract the second
result from the first for the other.
Notes : 1. Tlie preceding method is a direct consequence of Art. 26.
2. The above method finds a practical application when it is necessary to
find the difference l)etween the squares of two numerical numbers. Thus,
(235)«  '219)2 = (235 + 219) (235  219) = 454 X 16 = 7264.
Exercise 50.
Factor the following :
1. a^x^lP f ; 10 .r2  ^y2 . 25 a^x^  49 JV
2. a,*  ?/ ; 2r4  81 / ; .x^  ?/ ; x^^  f.
3. a86*81a^?/«; l100aH*c2; 16 a^^  9 6«.
4. 9 «2'.  4 a^ ; i «2  J /,2 . .^1 __ yl^
9
130 ELEMENTS OF ALGEBRA.
5. x^  /; (^ + hf  (c + df {X  yf  a\
6. a^[xyf'Axy^a hf I) {a ■\ hf  {a  hf.
7. (a + l)2(&+l)2;Xa+ir— C^^l)2; (753)2 (253)2.
8. (24 X + yf {2Sx yf; (1811)2 _ (689)2.
9. {5x 2)2 (x 4)2; (1639)2  (269)2.
10. a^" 1; 729x'^y xy^; aH  b^ ; a  h.
11. (2x+ a Sy (32xf; 64:X^  729 y^
12. (575)2  (425)2 ; 2 a  4 2:2 ; 25 a"  3 6''" ; a:^  3/6
59. Compound expressions can often be expressed as the differ
ence of two equal even powers of two numbers, and then factored by
the foregoing principles. In many such expressions it will be neces
sary to rearrange, group, and factor the terms separately. Thus,
Example L Factor x^  y"^ i a^  b^ + 2 ax  2h y.
Process.
*2_y«a26H2ar2&i/=:a:2+ 2 ax + a^  ¥  2hy  y^
= (x2 + 2 a a: + a2)  (62 + 2 6 y + t/2)
= (X + a)2 (b + yy
= [(X + a) + (^ + b)] [(x + a)(h + y)]
= [a + b i X + y][a — b + X  y'\
Explanation. Rearranging and grouping the terms, in order to
form the difference of two perfect squares, we have the third expres
sion. Factoring the third expression gives the fourth expression.
The square root of each term of the fourth expression is (x + a)
and (y + b). Adding these results for the first factor, we have
a + b + X + y. Subtracting the second result from the first, we
have a — b + X — y, for the second factor.
FACTORING. 131
Example 2 Factor 2xy + I  x^ — y^.
Process. 2zy + I  x"^  y* = I  x^ + 2 xy  y^
= 1 (x2 2x1/42/2)
= 1 (xyY Art 55.
= [l + (^2/)][l(^l/)]
= [l + xy][lx + yl
Example 3 Factor 4a^b^i4c^d^^8ahcd(a^\^b^c^(P)''
Process. 4 a* />» + 4 c2 </« + 8 a 6 c </  (a^ + />'  c^  d^y
= 4a^b'^ \8abcd + 4 c^d^  (a« + b^  c^  d'^y
= (2a6 + 2crf)2 (a2 I 62  c2  rf2)a
= [(2a6 + 2c</) + (a2f6*c2~rf''')][(2aft+2c(/)(a2 + 62c2f7i)]
= [2a6 f 2cd + a* + fc'*  c*  «/2J [2a6 + 2c</  a^  &« + c^ + d^]
= [(a2 + 2a6 + 6')(c^2crf + rf2)][(c2+2crf + f/2)_(rt2_2a6+/^'^)J
= [(a + ^y  (c  rf)2] [(c + dy  {a  6)2]
= [(a+6) + (crf)] [(a+6)  (c^)] [(c+rf) + (a6)] [(c+d)(a6)]
= [a + 6 + c  (/] [a + 6  c + dj [a  6 } c + </] [6 + c + rf  a].
Explanation. Arranging and factoring the first three terms, we
have the third expres.sion. The f*quare root of each term of the
third expression is 2ab + 2cd and a"^ + 1^ — c^ — d^ Adding and
subtracting these results, respectively, gives the fifth expreasion.
Rearranging (2a 6 and 2 erf suggest the proper anangenient) and
grouping the.se terms, gives the sixth expression. Factoring the
terms of the sixth expression, we have the seventh expression
Finally, factoring the terms of the seventh expression, we obtain
the result
Exercise 51.
Factor the following expressions :
1. a262c22&c; a^ { y'^ V^ 2 a y • l^a^V^^2ah.
2. 25a:2_2^_(56c_9c2; 2?Jf.2ax\ a^ y^ 2yzz\
3. 4a;2_i2a:y+ 97/281; T^^x^Uf, 4r*l+r)r9z2.
4. 16^* r2f   J 9rt2_6^4la:2_8a;y16y2.
132 ELEMENTS OF ALGEBRA.
5. x^ — li^ + '^'^^ — n^— 2mx — 2ny\ a* + b"^ — c^ — cf^
6. 12 xrj 4 x^ ^J y^ + z^; A x  1  4= x^ + 4 a\
7. (^2 _ 2/2 _ ^2)2 _ 4 2/2;22 . (^2« ^ ^2" _ c4m)2 _ 4 ^,2«52«
8. 4 ^2 _ 12 a a;  62  fZ2  2 c c^ + 9 a2 . 4 a:^ + 9 ?/2
16^2 25 6^2  12^?/  40 ^^s; Ax^  W' 2hcc\
9. :i'4  25 «« + 8 a2a;2  9 + 30 a^ + 16 a* ; 2/2 + 6 6aj
 9 62^2 _ 10 ^) ?/  1 + 25 ^>2 ; (a4«  4 a2"  6)2  36.
10. rz;2« _ 9 ^2 + ^2 _ 2 ^"^"^ _ 6 a &  ^>2.
11. a;6"4?/*'" + 12?/2'"^ + 2a3:i.^«_ 9:^2 4. ^6_
12. 4 ^2 _ 9 2/2 + 16 ^2 _ 36 ^2 _ 16 ^ ^ + 36 ny.
13. tt2« f ^>2« _ 2 a*^^)'*  c2'» ^ Aj^"*  2 c"* A;2m^
14. 4 ^2 + 9 ^.2 _ 16 (2/2 + 4 ^2) _ 4 (16 2/;2 _ 3 «^.).
15. a2_,_^5_9^,2 4.i^2. a^a292a2^,2 + ^4_^6a.
60. The method for factoring a trinomial consisting of two trino
mial factors depends upon the following axiom :
5. If the same numher he both added to and subtracted
from another^ the value of the latter will not be changed.
Example 1. Factor x^ + a^x^ i a*.
Solution. Adding and subtracting a^x"^, we have x* + 2a^x^ia*
 a^x^. Factoring the first three terms of this expression, we get,
{x^ + a2)2 — a^x^. Here we have the difference of two equal even
powers of two expressions, and it is equal to the product of the
sum and difference of their square roots. Hence, the factors are
a* + a a: + a:^ and a^ — ax + x^
FACTORING. 133
Process.
= X* i 2a2x2 + a*a«z«
= {x^ + a^ya^x*
= {xHa^a x) (xHa'^a x), or (aHa x+x«) (a«a z+z«).
Example 2. Factor 16 a*  17 a" 6^ + 6*. \
Process. I6a*na^b^^b* = \6a*n a^b^+9a^b^hh*9a^b^
= 16a<8a«fc2 + M9a262
= (4 a^ 62)2 (3 a 6)^
= (4a2 + 3a66«)(4a23a6Z>2)
= (rt + 6) (4 a  6) (a  ^) (4 a  6)v
Explanation Adding and subtracting 9 a* 6* to the expression
(to form a perfect square), arranging and factoring the terms, we
have the fourth expression (the difference of two equal even powers).
Factoring the fourth expression, we get the fifth expression. The
factors of 4 a' + 3 a 6  6* are a + A and 4a  b. The factors of
4 a* — 3 a ^ — ^^ are a — b and 4 a + b. Hence,
When a Trinomial is the Prodnct of Two Trinomial Factors.
Make the trinomial a perfect scjuare by adding the requisite expres
sion. Also indicate the subtraction of the same expression. The
resulting expre8.sion will be the difference of two squares. Take
the sum of their scjuare roots for one factor, and their diflerence for
the other.
Exercise 62.
Factor the following expressions :
1. 9a* + 3a2624.4fe*; aHOa^ + Sl; 16 X* \ 4 aP f ■\ 1/*.
2. a^ + a^iZ^f; Sla^2Sa^a^\Ua^; mHm^wa + w*.
3. 40^+8^*2/^+9?/*; a8 + a*fe2 + 54. 81a* + 36 aHie.
•1 25a^9a^}^\'l6b^; x^ + xyhy^; a^ + a^f + f.
5. 16a8 + 8a*&3+92>«; 9a*'{'38aH^+49b^; p^ + pHh
134 ELEMENTS OF ALGEBRA.
6. 49 a^ + 110 a?lP' + 81 &^ 9 ^* + 21 ^2 ^^ + 25 /.
7. m*" + m2» + 1 ; ^*" + 16 :2;2« + 256.
8. a23a&+&2; «4«_6^2nj2m_^j4m. 25m444mV+16?i*.
61. Frequently the terms of an expression can be grouped so as
to show a common factor. Thus,
Example 1 . Factor 2am + ^hm — cm — A an— Qhn + 2 en.
Process. 2am + 36mcm — 4an — 66n + 2cn
= (2 a m — 4 a 7i) 4 (3 & m — 6 6 ?i) — (cm — 2cn)
= 2a (m— 2n) h Sb(m — 2n) — c (m — 2n)
= (m  2 n) (2 a + 3 6  c).
Explanation. Grouping the terms of the given expression in
pairs ; taking the common factor 2 a out of the first, 3 b out of the
second, and c out of the third, we have the third expression. Divid
ing the third expression by m — 2n (the common factor), we have
2 a 4 3 & — c. Hence, the factors are m — 2n and 2 a + 3 6 — c.
Example 2. Factor 12 a^  4 a^b  S a x^ + b x^.
Process.
12a^ 4 a^b 3 ax^ + bx^ = (12 a^Sax^)  (4a^bb x^)
= 3 a (4a'^  x^)  h (4 a^  x^)
= (4 a2  x^) (3ab)
= l2a + x)(2a x) (3 a b).
Explanation. Grouping the terms in pairs ; taking the factor
3 a out of the first, and b out of the second, we get the third expres
sion. Dividing this by 4 a^ — x^, we have 3a — b. The factors of
4a^— x^ are 2 a + x and 2a — x. Hence, the factors of the poly
nomial are 2a + x, 2a — x, and 3 a — b.
Example 3. Factor 2mn — 2nx — my + xy { 2 n^ — ny.
Process. 2mn — 2nx — my + xy + 2n^ — ny
= (2mn — 2nx + 2 n^) — (my — xy + 7iy) ,
= 2n (m — X + n) — y (m — X + n)
= (m — X + n) (2 n — y).
FACTORING. 136
Example 4. Factor 4ax + 4x^ + 4ay\4y^8xy.
Process. —4ax^4x^4ay+4y^6xy = 4[ax+x^\ay+y^2xy]
= 4[{x*2xy{y^){axay)]
= 4[{xy){xy)a{xy)]
= 4(xy)[(x^y)a]
= 4{xy)[xya].
Example 5. Factor 2am^2an^2am2an\2a^2a^4a^n.
Solution. Remonng the common factor 2 a, we have w* — n*
7n — n + a — a*42an. Arrange the terms of this expression into
the groups m* — (n* — 2 a n + a*), and — (m + n — a). The factors
of the first group are m + n — a and m — n + a. Hence, m^ — n'
— m — n + a — a22an = m*— (n* — 2an4a*) — (m + n — a)
= (m + « — a) (m — n + a) — (m + n — a). Dividing this expres
sion by the common factor, m + n — a, we have m — n + a — 1.
Hence, the factors of the polynomial are 2 a, m + n — a^ and
mn\a 1. Therefore, 2am^2an^2am2an'{2a^
~ 2 a* h 4 a^ n = 2 a (m + n  a) (m  n + a — I).
Process. 2 am^  2 a n^  2 am  2 an + 2 a^  2 a* + 4 a^n
= 2 a[m^  n^  m  n + a  a^ + 2 an]
= 2a [(my  (n  a)^ (m + n a)]
= 2a[(m + n  a) (m — n i a) — (m + n  a)]
=: 2 a (m f n — a) [m — n + « — 1] . Hence,
To Factor a Polynomial by Gronping its Terms. Group the
terms of the polynomial so that each group shall contain the same
compound factor. Factor each group and divide the result by the
compound factor. The divisor and quotient will be the required
factors. If the polynomial has a common simple factor, remove it
first.
Note. It is immaterial what terms are taken for the different groups so that
each group contains a common factor. Tf the groups are suitably chosen the
result will always be the same, although the order of the factors may be
changed. Thus, in Example 3, by a different grouping of the terms, we have
2mn — 2nxmy + xi/ + 2n^ — ny
= (2mn — my) — (2nz — xy){(2n* — ny)
= w (2 n  y)  X (2 n  y) + n (2n  y)
= (2ny)(mx + n).
136 ELEMENTS OF ALGEBRA.
Exercise 63.
Factor the following :
1. a^ ■{ ab \ ac + be; a^c^ + acd — '2abc — 2bd.
2. am—bm—an+bn; 4:ax—ay—4:bx+by; af^+a^{cr\a.
3. Qax — Sbx — Qay + 3by; pr \ qr — 2^ s — qs.
4. ax— 2hx\2by\4tGy — 4:cx— ay.
5. 5a2 5&2_2a + 2&; ^ x^ + Z xy  2 ax  ay.
6. 2x^3^^4cX2; a'^x^a^x^a'^x^+1; mx2my
— nx + 2 7iy\ 4: X — a X \ 4: a — a^.
7. x^ + mxy—4:xy — 4:my'^\ 4:a^ + 4:x'^ + 5a — 5x — Sax.
8. 3a'^—Sac — ab + bc; a'^ x + ab x + a c + ab y + I'^y { b c.
9. ^ ax^ + 3 a xy—5bxy—'Sby'^; mn + np — mp — n^.
10. rii^ + np — mf — 7i?\ V^y^—2 x^y + ?>a^—21 xy^.
11. 21 a 5c + 3 «c 2&C 14 & 35; i^2_5^2/
+ 62/^+32: — 6?/; 2:^ — r?;22;_l,
62. Example L Factor a;^ + 1/2 + g'J  2 a; 1/ + 2 a; 2  2 ?/ 2.
Solution. The expression consists of three squares and three
double products. Hence, it is the square of a trinomial which has x,
y, and z for its terms. Since the sign of 2 x 2 is + , and 2xyis—,
X and z have like signs, while x and y have unlike signs. Hence,
one of the two equal factors is x — y + z.
.. x^ + y^ + z"^  2 xy + 2 xz  2yz= (x  y + z)^.
Example 2. Factor x^  3 x^ y \ 3 x y^  y^.
Solution. It is seen at a glance that the given polynomial fulfils
the laws stated in Art. 29. Therefore, one of the three equal factors
is x  y. .'. x^ — 3 x^ y + 3 X y^ — y^ = (x — yy. Hence,
FACTORING. 137
When a Polynomial is a Perfect Power of an Expression.
By observing the exponents, coetticients, and signs of the terms, find
such expression, as raised to a given power, will produce the polyno
mial. This expression will be one of the equal factors.
Exercise 64.
Factor the following :
1. a^ f 2 a 5 + Z>2 + 2 a 6 + 2 6 c + c2
2. a22a6 + 62_2«ci26c + c2.
3. a^+h^\c^\2ab2ac2bc; 16f 32a: + 24ic2
4. a^15a*x + 90a^x^ 2433:^  270a^JT'^+ 405aa^.
5. a^2ab + }r^+2ac + c^2ad2bc + (l^2cdi2bd.
6. 27x^i/l08a?x^y^64:a^+lUa^xy,
7. m^—2p x—2 n x\n^+]f—2 mn+2m x\3?—2 mp+2 np.
Miscellaneous Exercise 55.
Factor the following :
Vott. If the expression has a common simple factor, it should be first
removed.
1. 10ar»"30a:"40; ar^ + A^+l; \22^ifZ(SxyA8.
2. a;2 _ .56 a; + .03 ; a2 + f I a + 1 ; ^ ^ x  x^.
3. 3m«n83m*7i; 16^82; a^Sl; 6a;5448aJ*+72a^.
4. ar»ya5^^y^j; aH2^a3J^; 9(a + J)2
+ ^xy{a + byx^f.
138 ELEMENTS OF ALGEBRA.
5. a'^—2ax—4:a + x^+4:x; 8— 2x^—4:0^— 2x^; da^+a.
6. x^ + if ^^  3%; ^2« + 16 a" + 63; i a^'" +
(f ^t"  ¥ ^^") a^'"  6 a** +^.
7. m^ — a m — 71 m + a 71 ; a^ + 7 a — 8 ; 4 a^ — 4 2?^
2a + 2b; 49 a^H^"" + 7 (:i;2 + 32/^)a"53«^i _^ 3iz;i2/i
8. 2045aa2; ^8"  L2_8 ^4n ^ 1^
9. m^ — n^ — mp — np ; a;^ — ic^ + a;* — 2^^ ; a^ — a^ &^.
10. 380^^2. 8 ^10". _ 9 ^5m _. 1 . (^^ _ ^)4«
11. 6a.^2_2,__ 77. 12 2:2+108^+168; ^2^2^^^/
+ 2/2 — 5^—5?/; I 2^2 — Jg (5 m 71 + 3 2/) ^ + ^z'^^^'^ V
12. lTiy^Tfy^^ 2 2^2+ 5 ^?/ 3?/24aa; + 2a7/.
13. a2« + (^ _j_ ^) ^«^« + ^/2" ^7/; :r2 + (a + &) ^  2 a2
 a & ; a; 12  7/^ ;. 81 x^  22 2^2 ^2 + y^^
14. al2& + Z;13. ^3 + ^3_^3^^(^__^). a:4'»+c(a + 5)2:2«
— ah {a — c) (b + c) ] m^ + 71^ + 771 + 71.
15. a2_2,2_c24.^/2_2(^^_2,c); 4 + 4^'+2«y+^2_^2_2^2^
16. x^V{a + 2h)x\ah^h'^; x^^''+(ab)x^''2a'^2ah.
17. 250 (m nf±2; 8 (771 + nf ± (2 m  n^.
18. 52»c2»2;2'»_ 6"c"a2x" &~C"2:"+ft2; 49^4_i5^2^2
+ 121 2* ; (7?i + nf ± (m.  nf.
19. 4 (771 •— n)^ — (m — n); {m {■ Tif ± m {m + n).
FACTORING. 139
20. x^b{ac)xac{a\b)(b + c); 64:m^{12Sm^n^
21. 6 2^ + rS3^i/ + 6xf 6x^fxi/12f.
22. 2^'* {She + ac + ab)x'' i 'Sabc(b ■{■ c).
23. m3 + 4 m /i2 ± 8 n^ ± 2 m^ n ; {711 + 3 n)2  9 (m pf.
24. x^'' + (a \ b c)x^^'' ac  be; {x ^ yf  x
 y  6 ; 25 ^ + 24 x^i/ + 16 f,
25. 9a:*y*— 3 2^^—60,2 y^; m^— mri — 6 71^^47/1 ip 12 7^.
26. rrAn^{abzfm^n^{xy^2zf\ 81a2"_i99a»fe'»
+ 121 62; 81 a^"  99 a2«^,4« 4. 25 fts™
27. 18 a:2 _ 24 xy + 8 / ± 36 a; ^ 24 v/ ; 2 m2 j^ 2mn
 12 ?i2  12 am  36 a 71 ; a^a^ ± 64 a?iA
28. 2^^+ 3x'y282/*+ 28y + 4a:; 2y  ^ay \ 4.bx
\ G a X — 2x —4by.
29. 7W* 71 — 7?l2?i3 — 7;i3^2 __ ^^^ ,^4. ^j^4 _ ^^^^ ^ ^^4
30. 15 a^J  16 7/2  15 a a:  8 a; y + 20 a y ; (a  6)
(a2 _ c2)  (a  c) (a2  ^2).
31. c«^c2a2c3rf3 4.rt2; m87l3±512;247n,27l23077l7l3
— 36 71*; aa^ — 3bxy — axy\3b/.
32. 7m2 7n7i 69124. 16m 327i; 4 a;^ + 4 ajS  a;2  ar*.
33. 4 m8  4 7i3 _ 3 71 (712 _ m^) + 2 7?i (71  7?i)2
34 9m»±9a2m7; a.^~16y2 + a;±42^: (x2xy)^
— (a; — 2 a; y) — 6. Query. How many factors in the first part ?
140 ELEMENTS OF ALGEBRA.
35. 64.(4: x + yf  49 (2 a;  3 7/)2 ; (wt*  iii^  5)2  25.
36. m^ + m^ 7i + m 7i^ + iii^ ii^ + m^ n^ + rn? n^ ; (a; — y)^
 1 + 2:y (a;  ?/ + 1) ; (^2 + 4)216 a:2.
37. (m2 + 3 m)2  14 (m2 + 3 ??0 + 40 ; (m 7i  t!)^
 m ?i (tw % — 72. — 3) — 9.
38. :C2H _ ^n _ I ^ ^n ^ ^2n. ^f _ ^f^
39. 14^2 ^3_ 35a3 2;2+ 14a*:r; a;6 t/"!
40. 12:^:58^3^2+ 21 ^2^/; 64:cl± 27 ici
Separate into four factors :
41. {x2y)o(^{y2 x)t/; (^'"+ 6^"* + 7)2 (^'"+ 3)2.
42. 4 ^2 (^3 + 18 a &2) _ (32 a^ + 9 ^2 a?) ■ iQ ^^ ,^2
 (m2 + 4 ?^2  ^2)2 . (^4m _ 2 rt2^,2n _ J4«)2 _ 4 ^4m j4»
■ 43. x^ + ^./  8 0^6 ^3 _ g ^9 . ^9m _^ ^sm + 54 _^3»t + 64
44. m^ — 2 (?t2 + ^2^ ^.^2 _. (^2 _ ^2^2
Separate into five factors :
45. m'' — inP n^+ 2 7?i* n^ —m^n'^; 6 m* 7^2 + m^ n — 6 m^ n^
rn^n^l (a;2m _^ ^2n _ 20)2 _ (^^'^^n _ y2n _^ ]^2)2
46. ^7'« + ^4/n_;j^g^m__]^g. 16 ^7m_81 ^ »t_ ^g ^4m_^ 3]^
Separate into seven factors :
47, ^12m_^8mj4n_^4mj8n_^ J12n^
HIGHEST COMMON FACTOR 141
CHAPTER XII.
HIGHEST COMMON FACTOR.
63. The product of any of the factors of a number is a
factor of the given number.
Thu8, since 30 = 2 X 3 X 5, 6, 10, and 15 are factors of 30.
The product of the common factors of two or more num
bers must be a factor of each.
Thus, since 42 = 2 X 3 X 7, and 66 = 2 X 3 X 11, 2 X 3, or
6, is a factor of 42 and 66.
The product of the higliest powers of all the factors which
are common to two or more numbers must be the greatest
common factor of the given numbers.
Thus, since 24 = 2» X 3 and 36 = 2^ X 3^, 2^ X 3, or 12, is the
greatest common factor of 24 and 36.
The Highest Common Factor (H. C. F.) of two or more
algebraic expressions is the expression of highest degree
which will divide each of them exactly.
Thus, 3a:«y«i8the H.C.F. of 3a:«j/«, Qx^y\ and 15x<y«2.
Note 1. Two or more exprcRsions are said to be prime to each other when
they have no common factor. Thus, 5 a* and 9 b are prime to each other.
Example 1. Find the H. C. F. of 24 a« 6« c«, 60 a'^'c^y^
48a»62c«, and 36a2 6*c«x».
142 ELEMENTS OF ALGEBRA.
Process. 24 a^b^c^ = 2^ X 3 X a^ X b^ X c^ ;
60 a^b^c^f = 2^ X 3 X 5 X a^ x 6^ x c^ X 2^2 ;
48 a^b^c^ = 2* X 3 X a^ X b^ X c^ ;
36 aH^c^x^ = 22 X 32 X a2 X 6^ x c^ X x^
.'. theH.C.F. = 22 X 3 X rt2 X 62 X c2=: 12a2Z>2c2.
Explanation. Factoring each expression, it is seen that the only
factors common to each are 22, 3, a^, b\ and c^. Hence, all of these
expressions can be divided by any of these factors, or by their
product, and by no other expression.
Example 2. Find the H. C. F. of 2 x^  2 a: 3/2, 4 x^  Axy^, and
2 x*  2 a:2 3/2 + 2 x3 3/  2 X ?/3
Process. 2 a:^ — 2 x 3/2 = 2 x (x + y) {x — y) \
4 x^  4 xy* = 2^ X (x + y) {x  y) (a;2 + y^) ;
2 x^ 2x'^y^+ 2 x^ y 2xy^ = 2 x (x + y)^ (x  y);
.. the H. C. F. = 2 a: (a; + ?/) (a:  2/) = 2 a; (a:2 _ y^).
Explanation. Factoring each expression it is seen that the only
factors common to each are 2, x, x + y, and x — y. Hence, all of
these expressions can be divided by any of these factors, or by their
product, and by no other expression.
Note 2. If the expressions contain different powers of the same factor, the
H. C. F. must contain the highest power of the factor which is common to all
of the given expressions.
Example 3. Find the H. C. F. of 8a^ x^ + IQa^x^ + 8 a^ x\
2 a* a;2  4 a^ X  6 a6, 6 (a^ + a xy, and 24 (a2 a; + a x^y.
Process.
8a5x2+16a4x3+8a8x4= 23X a^Xx^ (a + xY;
2a^x^4a^x6a^= 2 X a'^X (a + a;) (3aa;);
6(a2+aa:)2= 2 X3Xa2x (a + x^;
24(a2x + aa:2)8= 2^ X 3 X a^ X x^ (a + a:)3.
The common factors are 2, a^, and a + x.
.. the H.C.F. = 2a2 (a + x). Hence,
HIGHEST COMMON FACTOR. 143
To Find the H. G. F. of Two or more Expressions that can
be Factored by Inspection. Separate the expressions into their
factors. Take the product of the common factors, giving to each fac
tor the highest power which is common to all the given expressions.
Exercise 56.
FindtheH.C.F. of:
4. 12 a^lJ^x^ and 18 a^bs^ ; 6 ci^xy, 8 aT^y, dindi ^tii^xy^.
5. loa^a^y^, ^a^x^f, and 21d^x^7/.
6. 12 2:8^2 22^ 18 a:*/^^, and 36 a:^^^^.
7. 20 c8a:V, 8a2x2yl, and \2a^x^yi.
8. a^hx ■\ al?x and a^h — l/^.
9. a^y'^ — z^ and ax^y ^h xy \ axz — hz.
10. 3a;*+8a:8+4ar*, ?^a^^ll3^+^3^, and Za^l^7?\2x^.
11. Za^x^yZa^xyZ^a^y and 3a2a:348 a'^ x
 3 a2 0^2 4. 48 a\
12. x^{x,{x^ 1)2 and 2^8 + 1 ; a^" + a:"  30 and
a^ _ a: _ 42; a:8 ^ 27, a:^ _ 9^ and 2 a^ + 5 a;  3.
13. a^ — a^y, a^ — xy^, and 7^ — xi^.
14. a;* f aj2 y2 _^ y4 ^nd a^ — 2ar^y+ 2xi^ — i^.
15. 12 (a  hf, 8 (a2  ^2)2^ and 20 (a*  ?>4).
16. 8 a: 2; (a; — y) (x — z) and 12yz(i/ ^ x)(y — z).
17. 4ar» 12a: + 9, 4a^ 9, and 4a^bx6a^b.
18. a?21f, x^^xy^^f, and 2 r* a;y~ 15/.
144 ELEMENTS OF ALGEBRA.
19. :r}' — if, (x^ — y^f, and ax^ — 1 axy + iS ay^.
20. ma^ — mx, 2x^\l^ x'^—2^ x, and 4:a?x^—4:a^x.
21. 24m7iC?/i+16 ??,4, 649i24, and IG^i^S^if 1.
22. a;2 + 4 a; + 4, ^3 + g, and 4 ic2 + 2 a;  12.
23. 16 ^3 _ 432, a;2  6 ^ + 9, and 5 x^  13 rr  6.
24. rii^—n?, m7i — 7i^+mp—np, and m^—m?n + mn^—n^.
25. 6 x^ — 9Q X, m a^ y — 8 m y, and 15 ^ a:^ — 60 ^.
26. a;6«ll2;3«+30, r^6«_i3^3«4.42^ a^i^j ^6n_^^3n_42.
27. a;3'' 125, ^2 « _ 10 ^« + 25, and 2 2^2« _ ^ ^n ^ 5^
28. Sa^'' 125, 4^2«_25, and 4a:2« 20^" + 25.
64. If the expressions cannot readily be factored by inspection,
we adopt a method analogous to that used in arithmetic for the great
est common divisor of two or more numbers. The method depends
on two principles :
1. A factor of any expression is a factor of any multiple
of that expression.
Thus, 4 is contained in 16, 4 times; it is evident that it is con
tained in 5 times 16, or 80, 5 times 4, or 20 times. In general,
Since a factor is a divisor, if a represent a factor of any expression,
m, so that a is contained in m, b times, it is evident that it is con
tained in r m, r times h, or r h times.
2. A common factor of any two expressions is a factor
of their sum and their difference, and also the sum and the
difference of any multiple of them.
HIGHEST COMMON FACTOR. 145
Thus, 4 is contained in 36, 9 times, and in 16, 4 timed. Hence, it
is contained in 36 + 16, 9 + 4, or 13 times, and in 36  16, 9— 4, or
5 times. Again, 4 is contained in 5 times 36, 5 times 9, or 45 times;
also, 4 is contained in 10 times 16, 10 times 4, or 40 times. Hence,
it is contained in 180 + 160, 45 + 40, or 85 times, and in 180160,
45  40, or 5 times. In general,
Let a be a factor of m and n, so that a is contained in m, b times,
and in n, c times. Then (m + 7i) + a = 6 + c ; also, (m — n) { a
= b  c. Again, since a is contained in m, b times, it is evident that
it is containe<l in r m, r times b, or rb times ; also, since a is con
tained in n, c times, it is contained in s n, « times c, or sc times.
Hence, rm ^ a=irb, and sn r a = sc. Adding these equations, we
have {rm ^ s n) ■Ta = rb\sc; subtracting the second equation
from the first, we have (rm — .<* n) + a = rb — sc. The last two
et^uations may be written (r 7n ± s n) ^ a = r b J^ s c. Therefore,
rm±sn contains the factor a. ,
Example 1. Find the H.C. F. of 4 a:«  3 a:^  24 a;  9 and
8 x»  2 or^  53 X  39.
Solution. The H.C. F. cannot be of higher degree than the first
expression. If the first expression divides 8 x*—2 x^ — b'Sx — 39, it is
the H.C. F. By trial, we find a remainder, 4 2:* — 5 a;  21. The
H.C. F. of the given expressions is also a divisor of 4 x* — 5 a: — 21,
because 4x* — 5x — 21 is the difference between 8 a:* — 2 z^ — 53a:
39 and 2 times 4 a:»  3 a;*  24 z  9 (Principle 2). Therefore,
the H.C. F. cannot be of higher degree than 4x' — 5z — 21. If
4 z^  5 a:  21 exactly divides 4 x«  3 r*  24 z  9, it will be the
H. C. F. By trial, we find a remainder, 2 z^  3 z  9. The
H.C. F. of 4 z2  5 z  21 and 4 z«  3 z*  24 z  9 is also a divi
sor of 2 z* — 3 z — 9, because 2 z* — 3 z — 9 is the difference between
4 z»  3 z»  24 z  9 and z + 1 times 4 z^  5 z  21 (Principle 2).
Therefore, the H. C.F. cannot be of higher degree than 2z* — 3z — 9.
If 2 z* — 3 z — 9 exactly divides 4 z'' — 5 z — 21, it will l)e the
H. C. F. By trial, we find a remainder, z  3. The H. C. F. of
2 z* — 3 z — 9 and 4 z* — 5 z — 21 is also a divisor of z — 3, because
z — 3 is the difference between 4z* — 5z — 21 and 2 times 2z* — 3z
— 9 (Principle 2). Therefore, the H.C. F. cannot be of higher de
gree than z — 3. If z — 3 exactly divides 2 x* — 3 z — 9. it will be
10
146 ELEMENTS OF ALGEBRA.
the H.C.F. By trial, we find that a: — 3 Ib an exact divisor of
2x^3x9. Therefore a;  3, or 3  a; is the H. C. F.
Process. 4x^3x^ 24a:  9 ) 8a;3 ~ 2a;2  53a;  39 ( 2
2 times the divisor, 8 a:^ — 6 x^ — 48 a; — 1 8
First remainder.
4a,2 5a: 21
4^2 _ 5
X times second divisor,
Second remainder,
X
2l)4a:83a:224a: 9(a:+l
4a:35a;22la:
2x2 2x 9
2 times third divisor.
2:
r23a:9)4x25a:2l(2
4x26a:~18
Third remainder.
X 3
2 a: times fourth divisor,
Fourth remainder,
3 times fourth divisor,
x3)2x2_3x9(2x + 3
2x2 6x
3x9
3x9
Therefore, the H. C. F. = x  3, or 3  x.
Note 1. The signs of all the terms of the remainder may be changed : for if
an expression A is divisible by — £, it is divisible by + B. Hence, in the
above example, the H. C. F. is a: — 3, or 3 — a;.
Example 2. Find the H.C. F. of 4x^  x^y  xy^  6 y^ and
7x8 + 4x2^/ + 4X^2_3^8
Process.
4x8— x2i/—x2/2— 5 2/8)7x8+ 4^2?/+ Axy^— 33/8(7
4 times first dividend, 28x8+16x2 2/+16x?/2— 12^/8
7 times first divisor, 28x8— 7x^y— 7xy^—Sby^
First remainder, 23x2^+23xp+23/ = 23 ?/(x2+xi/+.y2)
x2+xy+2/2)4x8— x^y— xy^—5y^(^4x 6y
Ax times second divisor, 4x8+4x23/+4x.?/''^
Second remainder, —bx'^y — bxy^—by^
— 5y times second divisor, —bx^y — bxy'^—b y^
Therefore, the H. C. F. = x2 + xi/ + ?/.
Explanation. Arrange according to descending powers of x, take
for the divisor the expression whose highest power has the smaller
coeflficient, and multiply the dividend by 4 (to avoid fractions).
Since 4 is not a factor of 4 x^ — x'^y — xy^ — 5
HIGHEST COMMON FACTOR. 147
given expressions is the H. C. F. of 4 x* — a;'* y  xy^ — b y* and 28 x«
+ 16 a^»y+ 16 a: y2 12 i/« (Principle 2). Since 23 y {x^ { x y \ y^)
is the difference between 4 times the dividend und 7 times the
divisor, the H. C. F. oi the given expressions is a divisor of it
(Principle 2). Therefore, the H.C. F. cannot be of higher degree
than 23y (x^ ^ ^y + y^) If the first remainder exactly divides the
first divisor, it will be the H.C.F. Since 23 y is not a factor of
the first divisor, it can be rejected. Therefore, x^ \ xy + y^ is the
H.C.F.
This method is used only to determine the compound
factor of the H. C. F. If the given expressions have
simple factors, they must first be separated from them,
and the H.C.F. of these must be reserved and multiplied
into the compound factor obtained. Thus,
Examples. Find the H.C.F. of 54x^y + 60x'^y^  I8x»y*
 132 a:V and 18x«ya  50 a^^y* + 2x*y* I2x^y\
Solution. Removing the simple factors 6x^y and 2x^y^j and
reserving their highest common factor, 2x*y, as forming a part of the
H.C.F., we are to detennine the compound factor of 9a:*  22x'^y^
3xy*\l0y* = A and 9x*  6x^y \ x^y^  25 y* = B. UA
exactly divides B, it is the H.C.F. of ^ and B. By trial, we find
the remainder y(6x*23x^y 3 x !/« + 35 .y«). The H. C. F. of
A and B is also a divisor of this remainder, because the remainder
is the difference between B and 1 times A (Principle 2). Reject y
from this remainder, since it is not a common factor of A and B, and
represent the result by D. The H.C.F. of Z) and 2 A (a multiple
ofi4) is the H.C.F. n\ A ami J5 (Principle 2). This cannot be of
higher degree than D ; and if D exactly divides 24, it is the H.C F.
By trial, we find a remainder, 153 y^ (3 x^  x y  5 y^). The
H.C.F. of D and 2^ is also a divi.sor of this remainder. Reject
153 y«, and represent the result by E. The H.C. F. of E and D is
the H.C.F. of D and 2 A ; and if E exactly divides D, it is the
H.C. F. By trial, we find that E is an exact divisor of D. There
fore, E is the H.C. F. of A and B. Hence, the H.C. F. of the given
expressions is 2x^y (3x^  xy  5 y^).
148 ELEMENTS OF ALGEBRA.
Process.
9x^22x^y^'3xy^+10y^ = A )9x*Qx^y+ x^y'^ 25y^=B{l
1 times the first divisor, 9x* 22xY3xy^+10y^
6 a;8 y+2S x'Y+'^ x 2/335 ?/'»
= y{6x^23x^yZxy^+36y^)
{3x + 2Sy
6x^23x'2y3xy^+25y^ = D) ISx*  44xY ^^y^+ ^Oi/* = 2A
3x times second divisor, l Sx*^9x^y 9x'^y'^+l0bxy^
Second remainder, 69^:^ 3bxY^^^xy^+ 20?/^
2 times second remainder, 138x3^ 1{)xY222xy^+ 40y*
23 y times second divisor, l 38x^y529xY~ G9xy^+S05y^
Third remainder, 459xV153x2/87652/4
= I53y^(3x^xy5y'^)
3x^xy5y^ = E)6xf^23x^y 3xy^ + 35y^ = D{2x7 y
2x times third divisor, 6x^~ 2x^y — 10xy^
Fonrth remainder, 2lx^y+ 7xy^\35y^
— 72/ times third divisor, —21x'^y+ 7xy^]35y^
Therefore, K.C.F. = 2x^y {3x^xy~5y'^).
Example 4. Find the H. C. F. of 90 x^ y^  200 x^ y^  10 x^ y*
and 144 x* 7/  64 a: i/*  16 x^ /  144 x^ y^.
Removing the simple factors lOx^y^ and 16 xy, and reserving
their highest common factor, 2xy, as forming part of the H. C. F.,
arranging according to descending powers of x, we have
Process. 9x^—xy^—20y^ ) 9a;8— Qx^i/— xy'^— 4y^{^
1 times the first divisor, 9^^ — xy^—20y^
First remainder, —9x^y +16^^ — —y(9x^—l6y^)
9x2162/2)9x3 xy^20y^{x
x times second divisor, 9x^—l6xy^
Second remainder, 1 5 x ?/2 — 20 i/3 = 5 ^/^ (3 x — 4 1/)
3x4?/)9x216y''=(3x + 4i/)(3x4 2/)(3x + 4i/
3x + 4y times third divisor, (3x + 4y)(3x — 4y)
.•. the H. C.F. = 2xy (3x — 4y). Hence, in general,
HIGHEST COMMON FACTOR. 149
To Find the H.C.F. of Two Polynomials that cannot readily
be Factored by Inspection, if the given expressions have simple
factors, remove them and iirraiige the resulting expressions acconling
to powers of a common letter. Take that expression which ia of
lower degree for the first divisor; or, if both are of the same degree,
that whose first term has the smaller coefficient. If there is a re
mainder, divide the first divisor by it, and continue to divide the last
divisor by the last remainder, until there is no remainder. The last
divisor will be their highest common factor. The highest common
factor of the simple factors multiplied by the last divisor will give
the H.C.F. sought. "
Notes : 2. If the first term of the dividend or of auy remainder is not ex
actly divisible by the first term of the divisor, that dividend or remainder must
be multiplied by such an expression as will make the fiist term thus divisible.
3. Observe that we may multiply or divide either of the polynomials, or any
of the remainders which occur in the course of the work, by any factor which
does not divide hoth of the polynomials, as such a factor can evidently form no
part of the H. C. F.
Exercise 57.
Find the highest common factor of :
1. ar3 4 2 ar2  13 a: + 10 and 2:3 _j. ,2 _ 10 a; + 8.
3. a^x^5xSsind 3^4x^11 x 6.
4 x*93^+29x^39x+lS and 4 a^ 27x^+56 a: 33.
5. 2^ — 5 ax^ 4 4 a^x and sd^ — ax^ + Za?x^ — 3 a^x,
6. 2fl0xf^%x^y and ^x'^^xif^Za^f^s^y.
7. 2a:Slla;2_9 and 4a;«+ ll2:*481.
8. 18 ar^ + 3 a:  6 and 18 a:^ _,. 95 ^ ^ 104 a; + 32.
150
ELEMENTS OF ALGEBRA.
9. 15 m^ n^ — 20 m^n? — 65 m^n — 30 m^ and 2amn^
\ 20 am n^ — 16 a mn— 186 a m.
10. 36 m^ + 9 m3  27?/i*  18 ?7i5 and 27^57^2  9 m%2
18 m* 72.2
11. S x^ — 3 X y + xy^ — y^ and 4:X^y — 5 xy^ + y'^.
12. mnx^ — 82 m 7i a; — 3 //i ?^ and m^ ti^ ^5 __ 28 ^^5 ^2^2
 9 m^n\
13. a;3  4 ^2 + 2 ^ + 3 and 2 ^*  9 ^3 ^ 12 ^2 _ 7
14. 16 x^^^xy + 10 7/2 and 6 :?:*  29 2;3 7/ + 43 i«2 ^2
20^?/^.
15. 2m^n — 10m^n7j'^+ IS m^ny^+ 224:mny^+2d4:ny^
and 4 m* 71 — 20 m2 ?i 7/2 — 48 in ny^ + 112 7i 7/*.
16. 27?2.'*a;4" 27?i"2;3"  4 7??,**^2« + 4^";:c" and 6m"a?5n
 18 m^'x'^'' + 12 77l"^3« _. 6 ^«^2n _ (3 ^n^n
Query. How many factors in this result 1
65. To Prove the Method for Finding the H. C. F. of any
Two Algebraic Expressions. Let A and B represent the ex
pressions, the degree of A being either
equal to or higher than that of B. Di
vide A by B, and let the quotient be m
and the remainder D ; divide B by D,
and let the quotient be n and the re
mainder E ; divide D l)y E, and let the
quotient be r and the remainder zero;
that is, E is supposed to be exactly con
tained in D.
We will first prove that B is a common factor of A and B.
From the nature of subtraction, the minuend is equal to the sub
trahend and remainder. Hence, A = mB \ D, B zz n D + E, and
Process.
B)A (m
mB
D)B{n
nD
E)D(r
rE
HIGHEST COMMON FACTOR. 151
D = r E. Since the division has terminated, E is a divisor of D.
E is also a divisor of n Z) (Principle 1) and of n D {■ E^ or B
(Principle 2). Hence, £ is a divisor oi mB (Principle 1), and of
mB + D, or A (Principle 2). Therefore, £ is a common factor
of A and B.
We must now show that E is the highest common factor.
Every divisor of A and B is also a divisor of m B (Principle 1),
and of A —mBy or 2) (Principle 2). Therefore, every divisor of
A and i5 is a divisor of nD (Principle 1), and of B — nD, or E
(Principle 2). But no divisor of E can be of higher dej^ree than E
itself. Therefore, E is the highest common factor of A and B.
66. Let i4, B, />, E, etc. represent any polynomials. Let m
represent the H. C. F. of A and B, n the H. C. F. of m and Z), and p
the H. C. F. of n and E^ etc. Evidently m is the product of all the
factors common to A and B ; also, n is the product of all the factors
common to m and Z), and /) is the product of all the factors common
to n and £, or /> is the product of all the factors common to A, By D,
and £, etc., which is their H.C.F. Hence, in general,
To Find the H. C. F. of Several Polynomials. Find the
H. C. F. of two of them; then of this result and one of the remaining
polynomials; and so on. The last result found ^vill be the H.C.F.
of the given polynomials.
Exercise 58.
Find theH.C.F. of :
4 4 y*, and 2 a;^ + 2 i/.
2. ai^ + a^Sa^^Jx9, 2^ x \ 2^ { s^\ 2x^ + 2x^,
and 3 + 3a:2_^^4.^^4^
3. m" a:3 4. 2 m" ar2 + m" J? + 2 m", 2 x'^ + 6 2^ + 4: s^,
Za^h93^+9x\6, '6x^ l23^Sx^6x, and Sx^
\2\5x\S2^.
152 ELEMENTS OF ALGEBRA.
4. 2x^5 x+ 6 3x\ 3 2;2 + 2 a;3  8 :i:  12, and
5. Sa^""  33 a;2" + 96 a;"  84, 68 x^"  92 x^""  24a;»
+ 32 x!^\ a;3" + 11 ^"66 :>j2«, 50 a;" + 20 r?;3«  60 r2;2H
 20, 5 x^"  10 a^" + 7 a;"  14, and 3 ^«"  35 a:^"
+ 162 2:^"  372 a;3« + 494 a;2»  192 x\
6. 9 a:2« + 4 X" + 2 2:3" _ 15^ 48 ^n + 30 _ 343 ^"
 24 x^"", 8 a;2« + 4 x^'' + 3 a;" + 20, and 2 x^"" + 12 a:^"
 94 a;" 60.
7. o 3^ — 2 X y^ — 5 x^ I/, 5 xy^— 6 y^ — 3 x^y"^ — x^y
+ x\ 9 a^  8 x^ y  2i) X7/, S xy^  7 x^y  2 y^ + 3x^
10 y^ — x^y^ — 5 x^y + S x!^ — 7 xy^, and x^ — x^y — x^y^
 x]^— 2y^.
Miscellaneous Exercise 59.
Note. When possible the student should separate the given expressions
into their factors by inspection.
rindtheH.C.r. of:
1. 7^ — xy^ and x^ + x'^y ■\ xy + y\
2. x^ — ?/2, {x — ?/)2, of — x'^y, and 2x'^ — 2xy.
3. 2 x^ — X — 1, X y — y, x'^ y — X y, and 3 x^ — x — 2.
4. rc6  6 X + 5, 2a^+ b8x + x\ x^ + x^  11 x
+ 9, and 42 2^2 + 30  72 x.
5. a;2_i8a;+45, 22:27a:+3, 2^29, and 33p7x6.
6. 6a;4" 3:r3n_^2n_^n_;^ ^^^ 3a^^  3a^''  2x^''
 cr"  1.
HIGHEST COMMON FACTOR. 153
7. a^  //3^ x^ + ^V 4 2/^ a:^  //, x^ + x// + i/,
a^ + sH^y — x^ — i/y and r^// 4 ar^y^ + .^3^.
8. 2ar»+2rt + 4aa;, x^+23^+2x+l, 7b+Ubx
+ 7 6^3+ 14 6r2, 3 :r2  (3 m + n  3) a:  3 m  n,
a^_ 2 2^2 4 1, and 2 2:2+ (2;? + ^ + 2) a: 4 2p + q.
9. a:*  27 &8 ic, (2:2 _ 3 5 ^)2^ a a^  a b x^  6 a b^ x,
and &2:*4 62a:3 4. 353^,
10. 4 2:4'»2 2:3«_^ 3a.»_9 ^nd 2 a:*" + 2:2'^  2 2:3»
+ 3 2:*  6.
11. (a + 6) (a  ^/), (« + 5) (6  a), and (ft + «)^(^^  ^)^.
12. 2 63 _ 10 a &2 _^ 8 a2 6, 4 a2 _ 5 « J 4. ^,2^ ^4 _ ^,4^
9a*3a68+3a252_9a8j^ and 3 a^ 3 a26 + aJ2_ j3
13. 3 2:3'*3m2.'2'» + 2m2a:''2m8 and 3a:3"+ 2m2a:*
+ 8 m3+ 12 ma:3«
14. (ni — n) {x — y), (m ^n)(j/ — x), and (n — m) {x—y).
15. 9 2r»+ 3 2:3+ 12a:+ 20 + 2:^ 3 2:2^2 + ^^ _^2 2r»
+ 12y2+4y + 8, G2:2^_a;6_.62^^32,4.24anda^»2^
+ 3 2:2y + 4 2:2 + 4 y2 + 12 y + 16.
16. a3+3a2'»6« + 3a'»62m_^^«^ Sa^^+oJ^'", 4a2"fe2'»
+ 12 a"fe3m + 8 ft*"", and a2 62«.
17. 2;* — ?n. 2:3 __ (^ _ 1) a;2 4. ^ 3. _ ^ ^j^^l a:* __ ^ ^
+ (w — 1) 2:2 ^ ,j 2; _ ^
18. 3n2ar»+ 12?;i27i2+ 3n3^ 15mn^x+ \2m^nx
— Ibmnx^ and 2 m 71 2:3 4. g ^^3 1^2 _ 2 n^ a:3 4. g ^3 ^ 2;
+ 2 m n^ x^ — ^ m? n^ X — 2 noi^ — ^ m^ nx^.
154 ELEMENTS OF ALGEBRA.
19. x'^ — ma^ — mi? x^ — iii^ x — 2 m^, a;^ — 6 ni^ + mx,
x"^ — 2 m^ — m x, 3 3^ — 7 m x"^ + S m^ x — 2 m^, and x^
— 8 7n'^ + 2 7?i X.
20. 12x^i/24.3^2/+ 12x^7/, {x^yxf'f, xy{x^ff,
and ^7^y^2^x^if^ 2^x^i/^fx.
21. a^ 2o?h  aV^+ 2h^, a^ + a^h  ah^  h\
«3 _ 3 a 62 + 2 63, a^ h^, 2>ac3hc + 2ah 2V\
a^b\ and 2h^ + a c  he  2 ab.
22. a2 _ (^ ^_ c)2^ (^ 4. ^)2 _ j2^ c2  {a + hf, and a^
+ 2a6 + 62+ 2 6c + 62 + 2ac.
23. a^e^ + a^s^Se, ^4'^+ 5 2:3«__ g ^n^ :i:6«_4 2,3"_96^
^3«_. 32:2« + 3a:"+2, a:^'^  9 2:2" + 20, ^nd 3a:3n^3^2.
+ 5 2f^ + 2.
24. ^  2 2^2 + 3 ^ _ 6 an(i ;z;4  a?3  ^  2 oj.
25. 4 ^ 2/^ ~ 2 2/3 + 6 ^2 ^ and 4 ^2 ^ ^ ^ x^ — A.xy^.
26. 35^447^2__i3^+X and 42^4+41;2^3_9^2_9^_l
27. m7z,3+2??z7i2+7'/i'/i+2m and 3?i^— 12 ?^3_ 37^2^5^^
28. 2m22/5+166m22/296m2 2/ + 108m2 and ^mTv^f
— 144 m 1^ y^—l^m 7? ?/2 — 108 m n^.
29. 2^4_6^.3^3^2_3^,+l and ^7_3^6+^_4^2+i2;r4.
30. 4a;H322;3+36^2^8^ and 8^6_24^4+24a;28.
31. a;^"— 82/3"»aj2«__2;"2/'» + 2/'" and ^2«__4aj«^+42/2m
LOWEST COMMON MULTIPLE. 155
CHAPTER XIII.
LOWEST COMMON MULTIPLE.
67. A Multiple of a uumber coutains all the factors of
the j^aven number with higJiest powers.
Thus, since 24 = 2» X 3, 2« X 3 is a multiple of 24.
A Common Multiple of two or more numbers contains all
the factors of the given numbers with highest powers.
Thus, since 12 = 2^ X 3 and 9 = 3^^, 2^ X 3^ is a common multi
ple of 12 and 0.
The Lowest Common Multiple (L. C. M.) of two or more
algebraic expressions is the expression of lowest degree
which can be exactly divided by each of them.
Thus, 6 a*x»y« is the L. C. M. of 6 a*, x y\ a:», and a«y«.
Example 1. Find the L. C. M. of 42 a^x y*, 56 a x*y^, 63 a«x*i/«,
and 21 a* x^y.
BolutioiL Separating the expressions into their factors, we have
42 a»a? y* = 2 X 3 X 7 X a*^ X x X y*,
56 a x*y^ = 2* X 7 X a X x* X y^,
63 a»x«y« = 3« X 7 X a» X x« X y»,
21 a*x»y = 3 X 7 X a* X x« X y.
2* X 3* X 7 is the least common multiple of the coefficierts 42, 66,
63, and 21 ; a* is the lowest power of a that can be evenly divided by
ich of the factors a*, a, a\ a* ; x* is the lowest power of x that can
Ihj evenly divided by each of the factors x, x*, x^, x^ ; y* is the lowest
power of y that can be evenly divided by each of the factors y*, y^, y*,
y. Hence, the L. C. M. = 2» X 3^ X 7 X a» X x« X y» = 504a«x*y«,
156 ELEMENTS OF ALGEBRA.
Example 2. Find the L.O.'M.. oi 6x^2 x, 9x^3 x, t5{z^+x y),
8 (x z/ + y'^y\ and \2 a^ x^ y^.
Solution. Separating the given expressions into their factors, we
have
12 a^x^y' = 22 X 3 X a2 X a:3 X y^
8(^2^ + 3/2)2^23 X 2/2 X (a: + y)2,
6(a;2 + ar2/) = 2 X 3 X a; x (a; + 2/),
6a;22a;=2 x a; X (3 a; 1),
9a;23a:= 3 X a; X (3 a; 1).
23 X 3 is the least common multiple of the coefficients ; aP is the
lowest power of a that can be evenly divided by a^ ; ofi is the lowest
power of X that can be evenly divided by each of the factors x\ x, x, x.
Similarly 2/, (a? + 2/), and (3 a:  1), each affected with its highest ex
ponent, must be used as multipliers.
Therefore, the L. C. M. = 2^ X 3 X a2 x a;^ x 2/^ X (x+yf X (3 a; 1)
= 24 a^x^y^ (x + y)" (3 x  1).
Example 3. Find the L.C.M. of 4 aa;2 2/2 + n aa;2/2  3a t/^
a;8 + 6 a;2 + 9 a;, 3 x^ y^ + 7x'^y^6xy% and 24 aa;2 22aa:+ 4 a.
Process.
4aa;22/2+ llaa;2/23a2/2= a X y^(x + 3) X i^x1),
a;3 + 6a;2 + 9 a; = xX (x + 3)2,
3a:32/3 + 7a;22/36a;.?/3= x y^(x + 3)X (3a;2),
24aa;2 22aa; + 4a = 2a X (4a;l) (3a;2).
.. the L.C.M. = 2aa;iy3(a; + 3)2 (4 a; 1) (3 a; 2). Hence, in
general,
To Find the L. C. M. of Two or more Expressions that can
be Factored by Inspection. Separate the expressions into their
factors. Take the product of the factors affecting each with its
highest exponent.
Note. The L. C. M. of two or more prime expressions is their product.
Thus, the L.C.M. of
(i^ + ab + b^, aS + b% and a^ + b^ is (a^ + ab + b^) (a^ + 62) (^s + J8).
LOWEST COMMON MULTIPLE. 157
Exercise 60.
Find the L CM. of:
1. 4&3^i/, ^^aT^f, and 63r/^2«.
2. 24:111 71^3^, Z(Sm^n^2^, and 4871828.
3. Ua^l^c^, 9aHc2, and ^(Sah^d^.
4. 12m*n2 2/3, l^mnf, and 24m^?i3.
5. 12aa;8y*, a:'"^ — ?/, 2;— 2a:y + ?/, and rr^ + 2a;^ ^ 2/2.
6. m^ (a;^ — ^), 71^ (^x — y), and a:"* — y^.
7. 2a:(a; y), ixyix^ ?/), and 62:3/2(2.4. y)^
8. 2:2 + a:  20, 2^^  10 a: + 24, and 2:2 _ 2: _ 30.
9. 2^2+22:, 2:2 + 4a: + 4, 2:243^42, and a^ \ o x ^^ e,.
10. 2:* + a2 2:2 + a* and 2'* — a a:^ — ft^ a; + a*.
11. a:2 _ 3 2;  28, 2^2 + 2:  12, and a^  10 a: + 21.
12. 15 (2:2^ xy% 21(2^ ar/), and 35 {xy"^ + f).
13. ar2  1, ar^ + 1, and a^  i.
14. '6x^\ llx^ 0, 32:2 4. 3^, _j. 4^ g^^j 2r^ + 52 + 6.
15. 2^+{a^b)x\aby a^\(a\c)x+ac, and a^{(b\c)x{bc.
16. mx — my — nx + ny, (x — y)^, and Zm^n—?nnn^.
17. a:2 4. (rt 4 2,) 2, 4 a 5 and ar2 + (a  i) a:  a 6.
18. ar2  1, a^2 4 1, 2:4 4 1 ^nd a:^ + 1.
19. a:^ + ar*y + a;?/2 f ?/3^ ^i — x'^y { xy^ — if, and ^
\ x^y xff.
158 ELEMENTS OF ALGEBRA.
20. 6 aa^+7 a^x^S a^x, 3 a^ x^ + Ua^x^ a\
and 6x^ + 39ax + 45 a^.
21. x^ + 5 X + 4:, x^ + 2 X  8, and a:^ + 7 ic + 12.
22.* 12 x^  23 a:^ + 10 2/^, 4a:2 _ 9^^ _j_ 5^2^ ^nd 30^2
— 5xy + 2y^.
23. a'^4&2, a32a2Z^+4a&2_8j3^ and a^+2a^b
+ 4 a 62 + 8 53
24. a m + <z '/I + & /?i + & rt and ax + ay + hx^hy.
25. 8a;2_38a;2/+352/^ 4a:2_:i,z/_53/2 and 2x^5xy7y^
26. " 2^2 + ?/2, x^ — n:2 2/2 + 2/^ and 2:^ + y^.
27. 60 a;4 + 5 a;3 _ 5 ^2^ 60 n;2 ?/ + 32 a; 2/ + 4 y, and
40 a:^ 2/ ~ 2 a:2 2/ — 2 a;?/.
28. {a + 6)2  (c + ^)2, (a + c)2  (& + ^)2, and (a + df
 {b •+ c)2.
29. 2^2 {_ ^ ^ __ ^2^ 2^ _ ^ y _l_ y2^ and a:^ + a;2?/2 + 2/*.
30. 3 a^* + 26 a;3 + 35 a;2^ 6 a:2 + 38 a:  28, and 27 x^
+ 21x^30x.
31. 12 a:2n _^ 3 ^n _ ^g, ^^ ar^« + 30 a;2" + 12 x\ and
32^2n_4Q^„_ 28
32. « (??/ — n), b (n — m), and — c{m — n).
33. (a b)(b c), (b a)(b c), and (b a)(c b).
34. a(& — a:)(a;— ^), b{c—x){x—a), and c(a — a:) (a:— 6).
35. a:*^  2 aj2n _,. 1 ^nd a:*" + 4 x^'' + 6 a^" + 4 a:" + 1.
Result. a;«" + 2 ar^"  a;^"  4 rr^"  x^" + 2 a;" + 1.
LOWEST COMMON MULTIPLE. 159
68. If the expressious cannot be factored by inspection, find their
H.C. F., then proceed as before. Thus,
Example L Find the L. C. M. of 2 z< + a:«  20 x"  7 x + 24
and 2 X* + 3 z»  13 x2  7 X h 15.
Solution. The H.C.F. of the expressions (Art. 64) is x^\2x3.
Dividing each expression (for the other factor) by x* + 2 x — 3, we
have 2 X*  3 X — 8 and 2 x* — x  5. Hence,
2 x« + x»  20 x2  7 X h 24 = (x2 h 2 X  3) (2 x«  3 X  8),
2 X* + 3 x«  13 x2  7 X h 15 = (x* f 2 X  3) (2 x'^  x  5).
.. the L.C. M. = (x2 H 2 X  3) (2 x2  3x  8) (2 x2  X  5).
Example 2. Find the L.C.M. of x«  Sx^ f 19z h 12, x» 6x2
+ 11 X  6, and x«  9 x2 f 26 X  24.
Solution. The H.C.F. of the expressions (Art. 66) is x  3.
Dividing each of the expressions by x — 3, and factoring the quotients,
we have
x»8x«+19x12 = (x3)(x25x + 4) = (x 3) (x1) (x4),
x»6xHnx 6= (x3)(x23xf2) = (x3) (x1) (x2),
x«9x226x24= (x3)(x26xf8) = (x3) (x2)(x4).
Therefore, the L.C.M. = (x3)(xl)(x2)(x4)
= x*10x»+35x250x + 24. Hence,
To Find the L C M. of Two or more Polynomials that can
not readily be Factored by Inspection. Find the H.C.F. of the
^^iven polynomial!*, and divide each polynomial by it. Then find the
L. C. M. of their quotients, and multiply it by the H. C. F.
Exercise 61.
Find the L. CM. of:
3. x^{2.x3, :r?\Sx^r:S, and a:^ + 4 r^ + a:  6.
160 ELEMENTS OF ALGEBRA.
4. ic* — m aj3 — ni^ x^ — m^ x — 2m^ and 3 a;^ — 7 m cc^
+ 3 m^ 5:: — 2 m^.
+ 38^2/3+ 16^2/* 10 2/^
6. a?3  9 :c2 + 26 ^  24, a;3  10 a:2 + 31 2;  30, aud
aj3 112^2+ 38:^40.
7. ^4 2)3 4 a;2+ 16 2^24, 2^3_ 5^2 ^_ 8^4, and
a:2 + 2 2:  8.
8. 2)3 _^ ^.2 _ 10 a: 4 8, ic2 + 2 a:  8, a;2  3 a; + 2, and
ic2 1.
9. 6 2;3 + 15 0^2 _ 6 ^ 4. 9 and 9 a;3 4. G 2,2_ 51 ^ 4. 36^
10. 2 ar^  8 a;4 + 12 a;3  8 2:2 + 2 a:, 3 a;5  6 2^ + 3 re,
aud a::3_3^2__3^_l
11. a:* + 5 2^3 + 5 rc2  5 2:  6, a:^ + 6 a:2 + 11 a; + 6,
and a:^ + 4 2^2 + 2^ — 6.
12. 2 2)3 + 7 a;2 + 8 X + 3, 2 x^  2^2  4 2; + 3, 2 a:^
+ 3 2:4 + 2 x^ + 3 ^,2 4_ 2 a: + 3, and a?* 4 2^2 + 1.
69. To Prove the Method for finding the L.C.M. of any
Two or more Algebraic Expressions. Let A, B, D, E, etc.
represent the expressions, F represent their H. C. F., and M represent
their L. C. M. Also, let a, b, d, e, etc. represent the respective quo
tients when Af B, D, etc. are divided by F. Then,
A =aF, B = bF, D ^ d F, E = e F, etc. (1)
F is the product of all the factors common to A, B, Z>, etc. The
quotients a, b, d, e, etc. have no common factor. Hence, their
L.C. M. is a 6 rf . . . , etc. and the L. C. M. of aF, bF,dF, etc., or
their equals A, B, D, etc., is ah d ... F. Therefore, M — abde F,
etc.
LOWEST COMMON MULTIPLE. 161
70. Let A, B, D, Ej etc. represent any polynomials. Let A'
represent the L.C. M. of A and B, P the L.C. M. of N and D, and
R the L. C. M. of P and Ej etc. Evidently R is the expression of
lowest degree which can be divided by P and E exactly ; also, P is
the expression of lowest degree which is exactly divisible by A^ and
Z>, and N is the expression of lowest degree which is exactly divisible
by A and B. Therefore, R is the expression of lowest degree which
is exactly divisible by ^4, ^, 2>, and E, etc Hence,
To Find the L.G.M. of Several Polynomials. Find the
L. C. M. of two of theni; then of this result and one of the remaining
expressions; and so on.
71. Let A and B represent any two expressions. Let F repre
sent their H.C. F., and M represent their L.C. M. Also, let a and b
be the respective quotients when A and B are divided by F. Then
A = aF, B = bF, and M = ab F. Multiplying the first equations
together (Axiom 3, Art. 47), we have AxB = aFXbF=FXabF.
Therefore, substituting for abF its value M, A B  F M. Hence,
in general,
The Product of any Two Expressions is Equal to the
Product of their EOF. and L.C.M.
Miscellaneous Exercise 62.
Find the L. CM. of:
aZ^a^hah^h^ and a^  2 aH  a 6^ + 2h\
2. a:4« _ 10 2:2 4. 9^ ^n 4. 10 r^" 4 20 ar**  10 a:" 21,
and a;* + 4 a;8"  22 ar»"  4 af 4 21.
3. 2:3"4ar2"3r+ Oa:"^^"' lOyS" and a:«*f2a;2nym
11
162 ELEMENTS OF ALGEBRA.
4. s^ ^ Sx'^ + x^ + 3x^ + x + S, 2a^+6x^^2x6,
r" + 2x^ + a^+ 2x^ + x+2, and 2a^+3x^+2a^ + 3x'^
+ 2;:c+ 3.
5. X y — b X, X y — a y, i/^ ^ 3 h y \ 2 h^, x y ~ 2 h^,
xy — 2hx — ay + 2 ah, and xy — hx — ay \ ah.
6. a^" + ^^4" h"^ + a3» 52m _j_ ^2n 53m _^ ^n 54m ^ j5m^ ^nd
7. .:c2'*  4 a'^"*, .:c3'' + 2 a"' *2« + 4 a^'"^;" + 8 a^^ and
^3« _ 2 ^'« ^2» ^_ 4 ^2m ^« _ 3 ^3m
8. 27^"" + {2a3 &)^2n _ (2 52 + 3a&)a;" + 3 ^)3 and
2a;2" (3 62 c)^?'^ 3 6c.
9. ^ 2:r2_^ 4^_8^ ^3+2^2_4^_8^ a^3a^
4:X + 12, and ^^  3 a;*  20 a;^ + 60 aj2 + 64 a;  192.
10. x^''{ah)x^ah, x'^''(hc)x''hc, a^^'x^^'h^
 a;6»62 + 58^ and a)2»  (c  a)x''  ac.
Find the H. C. F. and L. C. M. of :
11. 3 o[^ — 7 x^y + 5xy^ — y^, x^ y + 3 xy^ — 3 0^—1^,
and 3 a^ + 5 x^y + xy^ — y^.
12. 6^'5 4.i52,4^_4^2^_;l^Q^2^2^y and9^y27a^22^
— 6 a2a;?/ + 18 <x^z/.
13. 6^3« + ^2n_5^n_2 and 6 2^" + 5 2;2~3.2J*2.
14. a^ ah + h^ a^ + ah + h\ a^ + a^ h'^ + 6^ a^ + j3^
a^ — 6^, and (c^ — h^f.
15. 2rc2^(6^_io6)a330a6 and 3 a;2 (9 a + 15 6) a;
+ 45 a 6.
LOWEST COMMON MULTIPLE. 163
16. a:3n_ 92«J ^ 26 X'*  24 and a:3»_ 122:2+ 47 ^•"CO.
17. (a.2 + 62)c + (62 + c2)x and {x^  W) c + {h^  c^) x.
18. (2 2^2 _ 3 ,,i2j y 4. (2 m2  3 y2) a; and (2 m2 4. 3^2)^
+ (2x2+3m2)y.
19. 2:3«^ 2a'"a:2n_^ a2'"a:" + 2 a^^ and a:^" _ 2 a'"aj2«
+ a2m^„_2a3'".
20. a;3 +3a^y + 3a;^^/J + 2/, oi^  x^y ^ xy^  2f,
sc^ \ xy \ y^y and «* + 2^2 ^2 _^ ^
21. 20^^x^h 25a:< + 5a:8~2:~l, and 25 a;*10 0:24.1,
22. 2:8* — yS* a^sy — i/*", y^(af — f)^, and a:2«
 2 a;*y" + 2^'".
Find the L.C.M. of:
23. a:*7a:8_72^_^43a.4.42 and ar*9a:8+9a:3
H 41 a;  42.
24. ir3+4a:a+6a;+9, «:«+«» 2 a; +12, and a:2_^_i2.
25. 4a:«42:*~29a^»21 and 4a:«+ 24a;* + 41 2:24.21.
26. 2a:*ll2:8+32r24.i0a; and 32:*142:«6a;24.5a.,
27. 2:3 _ 6 22 4. 11 a; _ e, a;3 _ a:2 _ 14 a; + 24, and
2:3 + .r2  17 2: + 15,
28. 32:^ + 52:84.525^52.^2 and 32^2:8+2:2_2._2.
29. 9a:*+18a:3_.^_92^4.4 and 6a:*+17a:«10a:+8.
30. 2 m^ + m2  m + 3 and 2 m8 + 5 m2 ~ w  6.
164 ELEMENTS OF ALGEBRA.
CHAPTER XIV.
ALGEBRAIC FRACTIONS.
72. The expression (a + &) f (m + n) may be written — 77 .
It is read the same in each case.
The second form is called a Fraction ; the dividend is the
Numerator, the divisor is the Denominator, and the two taken
together are called the Terms of the fraction, a, b, m, and n may
represent any numbers whatever. Hence,
An Algebraic Fraction is an indicated operation in divi
sion.
A Mixed Expression is one composed of entire and frac
tional parts ; as, n
m\
a
Note. The dividing line has the force of a symbol of aggregation, and the
sign before it is the sign of the fraction and belongs to its algebraic value.
73. Multiplying or dividing the divisor and the dividend by the
same number does not change the quotient. For, if we multiply the
dividend by any number, as m, the quotient will be increased m
times ; if we multiply the divisor by m, the quotient will be dimin
ished as many times. A similar method of reasoning may be applied
to the dividend and divisor.
A fraction is in its lowest terms when the numerator and
denominator have no common factor.
7a^hc
Example 1. Reduce ^^^ 3,3 to its lowest terms.
^o 0/ C
ALGEBRAIC FRACTIONS. 165
Solution. The H.C.F. of the numerator aud the denominator
7a^bc X 1
is la^bc. Factoring, we have 7^2^^ X4ac ' ^^J^^^"^8 ^^^ H.C. F.,
we have Since the terms are prime to each other the fraction
4ac
is in its lowest form.
6 a* + ax 15 x^
Example 2. Reduce ,r: 2 . ^u^^ — TTZz to its lowest terms.
15 a* + lis ax — 15 x^
6a^\ ax  15 x' _ (3 a + 5 x) (2 a  3 a;)
^^°*^®**' 15a2+ 16aa:15x2~ (3a45a;)(5a3x)
_ 2a Sx
 5a3x'
Explanation. Dividing the terms of the fraction by their H. C. F.,
we have This result is in its lowest terms, since the
5a — 3x
numerator and denominator have no common factor.
x*jx^y\xy* — y*
Example 3. Reduce 7 — ^ 5 . to its lowest terms.
XT — x*y — xy* — y*
x^ + x^y+xy'y* _ (x*  y*) + (x*y f x y*)
Process. ^^^y_^^_yA  (^ _ y4) _ (^^ ^ ^^.^
^ (x ^ + y«) (x« y^) + xy (x« + y«)
l^ + y«) (a:«  ya)  xy (x« + y^^
^ (x« + y«)[x«y« + xy]
(x2 + y«)[x2y2_^yj
x^ + xyy^
When the factors of the numerator and denominator cannot be
readily found by inspection, their H. C. F. may be found by the
method of Art. 64, and the fraction then reduced to its lowest terms.
Thus,
. T> , 4a»+ 12a«6 afta 15!»» . ,
Example 4. Reduce ^ . , ..^ «. . — i» — Trut to its lowest
6 a" + 13 a*o — 4 ao* — 15 6*
terras.
166 ELEMENTS OF ALGEBRA.
Solution. 6a^+r3a^b4ab^ 1568) 4a^+l2aH~ ab^l5b^(^2
3 times the numerator, 1 2 a^ +36 a^lSab^ —45 b^
2 times the denominator, I2a^j26a^b8ab^30b^
First remainder, lOa^b + bab^—lbb^
= 5b(2a^+abSb^).
2a2 + a&362)6a8+ 13a264a62_ i568(3a + 56
6as+ 3a2&9a62
10 a^b + 5 ab^ 15 b^
10 a^b{~ 5 ab^ 15 68
.'. the H. C.F. of the numerator and denominator is 2a2+a6— 36^.
Dividing each term of the fraction by 2 a^ + a 6 — 3 6^^ we have
4a8+ 12a26a62_i5 58 2a + 56
6a8+ 13a264a62 15 68 3a + 56
Hence,
To Eeduce a Fraction to its Lowest Terms. Divide both
terms by their H. C. F.
Exercise 63.
Keduce to lowest terms :
75ax^y^' ^m^a^y'^^ 4:X^ + 6 xy'
72 m^n^x^\ mn^{a^y^f ^ 2^ + Sx+ 1
24:'m'nix''' w?n{a^ — j^)' x^ — x — 2
6 m^ 11m 10 20 (a^  if) af"!/^"
6 w2 19 m +10' 5 2:2 +5 ^2/ + 5^/2' ^imyn+i
3 m^ + 23 yyi  36 3 772^ + ^m^n + ^Trv^Ti^
4 m2 + 33 m  27 ' m* + m^Ti  2 m27i2 *
^ in + Zmx x^ — {a \ b)x + ab
4 ml — 4 mt 2:2 ' a:2 + (c — ^t) 2: — « c '
(m + w)2 — a:2 ^.s _ 3 ^2y + 3 2; y^ _ ^yS
m a: + 71 a: — a:2' a:^ — x^y — xy^ \ y'^
ALGEBRAIC FRACTIONS. 167
cr^ — (y + mf ac — ad — he \ hd
x^ \ X y ■{ m X* a^ — b^
0.^4 (g + b)x+ ah 27a + a^
a^Jt\a\c)x+ ac' 18a6a2+2a8'
10.
11.
12.
(a _ 6)2 _ c2' (6 4 a;)^  {a + c)2
m^ — m* n — m n* + 71^ a a^ — 6 ic*"*"^
m* — m^ 71 — m^ ?i2  ??i ti^' a^hx — h^a^
a8+3aH+ 3a62+268' 48 ar* + 16 a:2 _ 15 '
7*. Example. Reduce :7^ :r— ^ — to a mixed
expression.
ProcesB.
2x  3y ) 4x«  16x2y + 29a;i/2  22y« ( 2z2  sxy f 7ya + 5^^^^4—
10x2y + 29x1/2
10xgyH5xy«
14xy^22y«
14xy«21y«
 .y» •
Explanation. Dividing the numerator by the denominator, we
have 2 «* — 6 xy + 7y'' for a quotient, and a remainder of — y«. As
— y* is not exactly divisible by 2x  3y, we indicate the division and
atld the result to 2 x* — 5 x y + 7 y*. Hence,
To Reduce a Fraction to an Entire or Mixed Expression.
Divide the numerator by the denominator.
168 ELEMENTS OF ALGEBRA.
Exercise 64.
Eeduce to an entire or mixed expression
1 + 22J 2x^x^ 9x^+14:
' r^^^J^' 2x^ xS
x^2x^ 6 a3  13 o2 + 6 a  6
z. 
3.
5.
x+ 1' 3a^2a + 1
x^ + ax^3 a^x S a^ ^^3 4. 2 a:^  12 a:  13
x2a ' a;2 + ^  12
x^ — 2 x^y^ + f/ x^\(m + n+l)x + mn+a
x^ + 2 xy + y^ ' x + n
6a^5x^+ Ixb a^S"  a:^"?/" + x^'y'^''  y^ ^
2x+l ' ^'"  2/"
75. Every expression may be considered as a fraction whose
denominator is unity. Thus, a =  ; a^b — c^ = .
2;2 _ y2 _ 5
Example. Reduce x \ y to fractional form.
^ xy
a:2  «2 _ 5 X + y x^  y^ — t)
Process. x\y — ~ ^ ^
{X
^V)
X 
X (rc
y
x'
ar2
1
X(x
5)
5
X —y ~ X — y'
Explanation. Writing the entire part in the form of a fraction
whose denominator is 1, and multiplying both terms of it by x — y,
we have the third expression. Since the sum or difference of the
quotients of two or more expressions divided by a common divisor,
is the same as the quotient of the sum or difference of the expressions
divided by the same divisor, we have the fourth expression. Uniting
like terms, we have the result. Hence,
ALGEBRAIC FRACTIONS. 169
To Reduce a Mixed Expression to the Form of a Fraction.
Multiply the entire puit liy the denouiinator ; to the product annex
the numerator; unite like terms and under the result write the
denominator.
Notes : 1. In the above example, since the sign before the dividing line
indicates subtraction, we most subtract the numerator, x^ — y^ — 5, from
{X + y) (x  y).
2. If the sign of the fraction is — , and the numerator is a polynomial, it
will be found convenient to enclose it in a symbol of aggregation before annex
ing it to the product.
Exercise 65.
Reduce to fractional forms :
1. a — X \ ; — ;  \ I] a + r .
a \ X m — n a — b
2. „,._,„, + ,._^^; ^^j;±_^^ (.,).
, « — m^n^ a . „ . . i m* — 1
3. m n H ; m^ + m^ + ?>i f 1
m n m — 1
4. ic + 1 4 ; ; 3^ — 3 + 1 ; w (a: + ?/) + — —
7/r — ar x r t
2 n (3 w2 + ?i2) wi* + w*
6. m + w ^ — , — ^2 — ; (w + ny — r^ •
7 a^m _a^,^«^ y8»_
a^m ^ sTir ■\ 7/2"
8. .^^,2^/+y»^'"'^^"'^°rr^'^"
170 ELEMENTS OF ALGEBRA.
76. It may be sliown by multiplication (Art. 22) that :
{+a)i+b) =ab; («)(&) = ab.
(+a)(+b){\c) =abc\ {a){b){+c) =abc.
(+a){+b){+c){+d) = abed; {a){b)(c){d) = abed, etc. Hence,
In an indicated product of any number of factors, all the signs of
any even number of factors may be changed without changing the value
of the product. Thus,
(xy) (:yz) = (yx)(zy);
(w x) (x y) (y~z) = {X  w) (x  y) (z  y), changing the
signs of the first and third factors.
Note. In order to multiply a product containing several factors by a given
expression the student must be careful to multiply only one factor of that
product by the expression. Thus, in order to multiply both terms of the
fraction ; — '— — ~ by a, we must multiply either a\b or c + d and
m \ n or X } y hy a.
77. It is often convenient to change the order and the signs of
the terms of the numerator or denominator, or both. Thus,
Change the order and the signs of the terms of the numerator and
denominator of the following fractions :
b — a m — n
1. • 2.
y X ' (jc — h) {x — m)
Solutions : 1. Multiplying both terms of the fraction by —1, we
have
b — a __ (b — a) X —I _a — b
y—x~{y~x) X —l~x—y'
2. Multiplying the factor x — m and the terms of the numerator
by — 1, we have
m — n _ (m — n)x— 1 n — m
(c ~b){x m) (c  b) [(x  m) X  1] (cb)(m~x)'
Multiplying the factor c — b and the numerator of this fraction by
— 1, and since adding a negative quotient is the same as subtracting
a positive quotient, we have
ALGEBRAIC FRACTIONS. 171
n — m _ (nm) X — 1 + (n  m)
(cb)(mx)~'^i(cb)Xl](mx)~~ (bc)(mx)'
Change to equivalent fractions having the letters arranged alpha
betically, and the first letter of each factor in the numerator and the
denominator, positive :
x — m (b — a){c — a)
3.
{ba)(a~c)(yx) " (d  a)(c  b)(n  m)
Solutions : 3. Multiplying the numerator and the factor y — x
by — 1, we have
X ~ m _ m — X
(b a) (a c) (yx)~ (b  a) (ti c)(xy)'
Multiplying the numerator and the factor b — a of this result by
— 1, we have
(6 a) (a c) (x  y) (a b) (a ^ c) (x  y)
4. Multiplying the factors c — a and n — m, b — a and c — 6 by
1, respectively, we have
(6 a)(c a) (a  b) (a  c)
{d  a) (c 6) (n  m) (d  a) (b c)(m n)
Q;„,n„rlv (ah)(ac) (a  b ) ( a  c)
oimuariy, ^^ _ ^^^ ^^ _ ^y ^^  n) ' (a  d) (b c)(mn)
(d — a) (c — o) (n — m) (a ^ d) (b — c) (m — n)
Exercise 66.
Change each of the following fractions to four equivalent ones
with respect to the signs of letters :
^ 7W — 71 _ a — Jj m VI \ n ^ a
a — b^ m + n — x' a — h \ x' m — n \ a*
172 ELEMENTS OF ALGEBRA.
Change the following fractions to equivalent ones having m and n
positive in both terms :
m — a a + m — X a + b — n
b — n ' b — m — y^ a — b + m
X — m X — m {a — m) (b — m)
y — n^ (y — m)(z — n)' (c — m) (x — n)(y — m) '
Change the following fractions to equivalent ones having the let
ters of the terms arranged alphabetically and the first letter of each
factor in the denominator positive :
. 2x?> — y 3 — c + ft
[m. — a) {2 X — b) (b + a)' (y — x) (m — n) (a — c)'
^ (x — m.)ba
5. r^ T77T
xy mn{c — b) (b — a) {c — a)
{jjx)yx
cb a {b — a) {z — y) (c — a) {y — x) (n — m)'
78. Fractions having a common denominator are similar.
Thus, ^, — =, and — ^ are similar.
ab ao ab
2x 3 5 n^
Example 1. Reduce r — 5, «, and . — i to similar fractions
having the lowest common denominator.
Solution. Evidently the lowest common denominator is 20 in^n^x^f
the L. C. M. of 5 m^, mn% and 4x^. Dividing 20 m^n^x^ by the
denominator of each fraction, and multiplying both terms of each
fraction by the quotient each by each, we have
ALGEBRAIC FRACTIONS. 173
2x 2x X 4n^x^ 'Sn^x* .
6 m« ~ 5 m* X 4 w» x« ~ 20 m^n*xfi'
_3 3 X 20ma:*_ _6()mx*_.
5na _ 5 n» X 5 m^ n» _ 25 m ^ n^
4x^~4x^ X 5 m^n* ~ 20 m^n^icfi*
Example 2. Reduce ^aJ^s^^^is * a:«!4x5 ^ ^'^^ x44^r + 3
to similar fractions with lowest common denominator.
Solution. The lowest common denominator is (x — 3) (x — 5)
(x + 1) (x + 3), the L. C. M. of the denominators. Dividing the
L. C. M. by the denominator of each fraction, and multiplying both
terms of each fraction by the quotient each by each, we have
x1 (a;l)X(a:+l)(x+3) (x + 3) (x^  1)
x^8x+15~ (x3)(x5) X(x+l)(x+3) ~ (x + l) (x5)(x29)'
x+3 (xf3) X (x3)(x+3) (x + 3)g(x  3)
xa4x5~ (x5)(x+l)X(x3)(x+3)~(x+l)(x5)(x29)'
x5 _ (x5)X(x5)(x3) (x  5)^ (x  3)
x«+4x+3 ~ (x+3) (x+ 1) X (x5) (x 3) ~ (x+ 1) (x5) (x^9)*
Hence, in general,
To Reduce Fractions to Equivalent Fractions having the
Lowest Common Denominator (L. CD.). Find the L.C.M. of
the denominators. Then multiply both terms of each fraction by
the quotient of the L. C. M. divided by the denominator of that
fraction.
Kotef : 1. When the denominators have no common factors, the multiplier
for both terms of each fraction will be the product of the denominators of all
the other fractions.
2. In all operations with fractions it is better to separate the denominators
into their factors at once; and sometimes it is also convenient to factor the
numerators.
3. It will he observed that the terms of each fraction are multiplied by an
expression which is obtained by dividing the L. C. D. by its own denominator.
It is not necessary to state how the multiplier is obtained in every expression.
174 ELEMENTS OF ALGEBRA.
Exercise 67.
Eeduce to similar fractions with L. C. D. :
a m X ahe 1 2 5
b' n' y^ mn^ ah' ac he'
m + n m — n n a a — n n
ah be a c b in da
m + 2 n 2 m. — 3 n 5 m — n
Sm ' 6 n ' 10 m 7^ *
1 x + 2 x2
6.
x'^V x^V x2' x^x2
m •— n m + 2n m^ 1
m + n' m — n m^ — 7i^' a + b' a — h' a^ + b'^
8m + 2 2ml 3m +2
ft
m — 2 ' 3 m — 6 ' 5 ??i — 10
_ xy m — n
8.
Tfix — rmy •\ nx — ny' 2 oc^ — 2xy
m n a
m + x' m^ + x^' m^ — mx + a^
X y m
x^ — xy + y^' x^ + X y + y^' ic* + x^y^ + y^
^^ x — y x^ + ?/2 y x^ + y^
11.
x^ + xy + y^' x^ — y^' x — y^ 5
12.
13.
ALGEBRAIC FRACTIONS. 175
b X
(a f xf  62' (5 _^ 2^)2 _ ^2» a:«  (a + 6)«
ate
(ca)(6c)' (ac)(c6)' {ca)(cl)
a _ g X 1 g
Sestioii. (^_^)(^_c)  f(ca) X l](6c)  (gc)(6c)* **
3a 4a 5 am n y
3"=^' o^' (a 3)2' n^' wrn:' r=^2
12 3 4
14.
15.
(2a;)(3aj)' (a:l)(2a:)' (a:2)(l^)' (a:l)(a>2)
1 1
Suggestion, (g  x) (3  a:) = (x  2) (x  3) = ^*^
16.
17.
18.
19.
(m — x) (a; — n) * (a: — ?w ) (a — a;) ' (a: — a) (n — a;)
1 + a: 2 i a;
(la;)(2a:)(a;5)' (x  1) (2  aj) (3  a:) (5  a;)
a; 3 a; 2 g2 + 4 2
4^=^' a^+a;6' 96a;4ar»' ir2a:6'
^ ar^"* + 1 A2m_ 1
a:*" 1' a^'" + 42:2.^_ 3' a^« + 2 ar»'"  3 '
79. Example 1. Find the sum of t , \^ and .
o a n
Solution. Multiplying the terms of the first fraction by dn,
of the second by bn, of the third by 6rf, and adding the results
(Arts. 32, 14), we have
a c m _adn hen hdm _adn\hcn + bdm
h'^ d'^ ~ii~ h(U'^ bd^'^ bdTk "^ hdik ^~ "
176 ELEMENTS OF ALGEBRA.
m a
Example 2. Subtract  from t
n
Solution. Multiplying the terms of the first fraction by 6, of the
second by n, and subtracting (Art. 19), we have
a m an hni an — hm
h n bn bn b n
2a3h ^ Sx2b
Example 3. Subtract — ^^ irom — ^ •
Solution. Keducing to similar fractions with L. C. D., we have
3 a: 26 2 a  3 6 _ 6ax  4ab 6ax9bx
3x 2a ~~ 6ax 6ax
6 ax — 4ab — (6 a x — 9 b x)
~" 6ax
■'J _ b(4:a9x)
~ Hax
T^. 11 p 2x — my Zx — ny
Example 4. Find the sum ot a I ' and b
m n
Solution. Uniting the entire parts, and reducing to similar
fractions, we have
/ 2x — my\ (^ 3 a!; — nwN , (2x—my)n C3x—ny)m
[a+ )+ [h ') = « + &+ ^^^ ^^
V m / \ ^ J '^^ ^^
(2x — my)n — (3x — ny)m,
= aib\ ' ■ '
mn
, (2 n  3 m) a;
mn
Note 1. If the sign of a fraction is — , care must be taken to change the
sign of each term in the numerator before combining it with the others. In
such case the beginner should enclose the numerator in parentheses, as shown
in the above work.
2^; g x+2 a;l
Exampi^e 5. Simplify ^2 + 3^ + 2 " x^2x3 " x^x6 '
Process.
ALGEBRAIC FRACTIONS. 177
lar6 xj 2 x \ I
x2 + 3x+2 x^2x3 x^x6
2 (x  3) x + 2 x+ 1
= (x + l)(x + 2) ~ (x + 1) (x  3) ~ (x 4 2) (x  3)
_ 2(x3)X(a?3) (xt2)X(x42) _ (x4l)X(a?+l
"(x+l)(x+2)X(x3) (x+l)(x3)X(x+2) (x + 2)(x3)X(x+l)
_ 2 (x  3)^  (x f 2y (x+ 1)3 _ 1318X
(x + 1) (x h 2) (x  3) ~ ix+ 1) (x + 2) (x  3) *
Notei : 2. In finding the value of an expression like — (x f 2)*, the be
ginner should first express the product in a parentheses and then remove the
parentheses as above.
3. Sometimes it is better not to reduce all the fractions to the L. C. D. at
once. Thus,
14 6 4 1
Example 6. • — 5 I rTT +
x — 2y x — y x x + y x + 2y
1 1 4 4 6
+ iTTzr  z —   zrr7. + z
x — 2y xh2y x — y x + y x
x + 2y x2y 4 (x + y) 4 (x  y) 6
(x2y)(x+2y)'^(x2y)(x2y) (xy){x\y) (x+y)(xy)"^x
2x 8x 6
~ x^  4 y2 x2  y3
_ 2 X (x^  y^) _ 8 X (x«  4 y«) 6
 (x« 4 f) (x^y^) " (xa  y2) (x3  4 y^ + X
_ 30xy''6x« 6
(x»4y«)(x«y2)'^x
_ (30xt/'6x»)x 6 (x2  4 y2) (x^ _ yS)
 (x2  4 y3) (x2 y*)x'^ X (x^  4 y^) (x^  y«)
24 y* TT .
= x(x«4y^(x»t/)  Hence, m general,
To Add or Subtract Fractions. Reduce to similar fractions
with L.C. D.; add or subtract the numerators, and divide the result
by their L. C. D.
12
178 ELEMENTS OF ALGEBRA.
Exercise 68.
Simplify :
2a — 5 S a — 11 b + c a + c a — b
12 a "^ l8 ' T^ "^ T6 97" '
X 2x ^x^^ xy xy^ xP"y^
^ m n fm + n a + b\ /m — n a — b\
' ab ac b c'' \ n a J \3?i ^a )'
3 + ^^ 4am a /5 4 3\ /I 2 3\
4. + + ^; +)+ )•
n an 6n \m n xj \m n xj
5ab 7a + Sb _ /2a a  b \
2b "^ 6^ \b "^ 3 & y *
6. (^,. + ^) + (3m^)(4m + ^).
^ a^—bc ac—b^ ab — c^ 2 a^—b^ b^—c^ c^—a^
i c ac ab ' a^ b^ (? '
8. (m + n ^ ) — ( 2m — 371 + — V
\ mxj \ nxj
\5 a; Zy 'omj \6 x 10 3/ 7 m/
a + & b — c c — a ab'^ — b(? — c 0?
11.
5 ab c
1 1 :r42 x2 3
oj— 5 ^—4' 2;— 2 a;+2' 2m(m— 1) 4m(?w— 2)"
ALGEBRAIC FRACTIONS.
179
12.
2a;/i — 3&71 2a7n + 'Sbn 1
1
'6vin{ia—7i) 3 7/1 7i (?/i + 7i) * ic^*4ic + 4
"x'^+xG*
13.
x + y J^ — y ' x—m X'\m x
14.
1 m+3 7 2 3
... 1 •> "T A 2 10> .J.^ "
2m3
1 A ™2 1 •
m n 2 mn 1 (a + 2xy^
m i 71 m — 71 ir? — r?' a — 2x a^ — S a^'
2: 1 1 11
16.
xy— if' x — y y' m^— (n + xj^ a^—(m + ?i)^
x^+'Sx^f+y* a^xy\ii^ 2 3? x + y
x^ — ^ X — y ' x*—y* x^+x?y+xif+y^'
,Q x + 4: , xhS x+2
3?+ ox{ 6 a? + 6 x+ 8 3^+7 x+ 12
1 mn m — 71
7n + 71 TTi^ \ n^ 7n^ — m 71 \ 71^ '
90 1 1 X X
7n + X m — X " (m + xf (tti — x)^'
21 __i L_ + ^ 4. _ ^
88ic 8 + 82:^4 + 40^2^ 2 + 2a;*
22 24a; 3 + 2a : 3  2 a ;
912a; + 4ar» 3  2 a: "^ 3 + 2a; '
00 « + & ?^ + r r + ^
(6  c) (c ~ a) ■*" (c  a) (a  ft) ^ (a  I) (h  c)
180 ELEMENTS OF ALGEBRA.
24 ^+^y , x^2y x^y
4.{x+y)(:y+2y) {x+y){xVZy) 4.{x+2 y) {x+Z y)'
^^ he ac , ah
(c — a) (a — b) (a — b) (b — c) (b — c){c — a)'
26 5(2^3) 7x _ 12(3a^ + l)
• 11(6^2+ ^1) 6 2;'^+7ic3 ll(4:X^ + Sx+3)'
x _ y x^y+xy^ x^ + 'f ^ — y^
x^+y^ Q^—i/ x^ — y^ ' a^—xy\y'^'~x^+xy\y'^'
28. ^" + ^* 1^
a;3 + ^2 _ 49 ^ _ 49 2:2 _ e ^ _ 7
29 ^ + ^' a; 4 & ^ + 0^
30.
{a — b) (a — c) (b — a) (b ~ c)
Suggestion. In finding the L. C. D. it is better to arrange the
letters alphabetically. Thus,
& a b a X —1
+ VI — wi — X = / — jaT N + m — \t:^ — TT7I — \ = etc.
{ab){ac) ^ {ba){bc) ~ (ah){ac) ^ [(ba) X l](bc)
x^+2x+4: oc^—2x\4: x2a 2{o?Aax) 3^
x\2 2—x ' x\a a^—x^ x—a
32 1 1 1 . 1 , ^
(m2)(x'+2)^(2m)(^ + m)' 2a;+l 2ajl
4a: 2 a: 3 a?2
+ . T—o\ — ^o ...... +
14^2' ^_l_4 :x:2_4^+i6'^^64
33. 7 iw T +
(a — &) (a — c) {b — a)ib — c) (c — a) {c — h)
ALGEBRAIC iRACTIONS. 181
Q C
80. Example 1. Find the product of r and ^ .
a , c
Solution. Let i = a:, and 3 = y. Multipljring both members
of the first equation by h and both members of the second by d (Art.
47, Axiom 3), we have a = bxy and c = d y. Multiplying these two
equations together, we have ac = bdxy. Dividing both members
of this equation by 6 d (Art. 47, Axiom 4), gives
ac _ a c
^ = xy. Buta:i/ = ^X^.
a c ac
Therefore, T ^ ^ = r, . Hence, in general, «
To Multiply a Fraction by a Fraction. Multiply the numera
tors together for the numerator of the product, and the denominators
together for the denominator of the product.
Notes : 1. Similarly, we may demonstrate the method when more than two
fractions are multiplied together; also, for fractions whose terms are negative,
integral, or fractional.
2. Since an entire or muted expression may be expressed in fractional form,
the method above is applicable to all casesl Thus,
^a m ^a am a / ,n\ a^/>».n\ am , an
r. o Ti'j.u J . 4x316x415 x26x7
Example 2. Fmd the product of ^ q , o — tt, tts — r= ^
,, „2 , ^ 2x21 3x+ 1' 2x2 17xf 21'
and
4 x*  20 X f 25
Process.
x26x7 4x«l
4x=»l
4 x2  20 X f 25
4x«16x15
2 x« h 3 X H 1
(2x3)(2x5)
"2xa17x+ 21 ^4x220xf25
(x7)(xhl) (2x+l)(2xl)
(2x l)(x 1) ^ (2x3)(x7) ^ (2x5)(2x5)
(2x  3) (2x  5) (x  7) (X H) (2x f 1 ) (2xl) _ 2xl
(2x l)(xf l)(2x 3)(x7)(2x5)(2x5)~ 2x6'
182 ELEMENTS OF ALGEBRA.
Explanation. Factoring the iiiuneiators and denominators of the
fractions, multiplying the numerators together for the numerator of
the product, and the denominators together for the denominator of
the product, we have the third expression. Reducing the third ex
pression to its lowest terms, gives the result.
Notes : 3. Observe the importance of factoring the terms of the fractions
first. Also, indicate the multiplication of the numerators and denominators,
and divide both terms of the fraction by their H. C. F. before performing the
multiplication.
4. If the factors are mixed expressions, sometimes it is better to change
them to fractional forms before performing the multiplication. Thus,
/ ah \ / _ ah \ _ a^ iA _ a^lfi
V^ ab)\ a + b)~ab a + b~ a^b^'
2 x^ \ 3 X 4 x^ Qx
Example 3. Find the product of — r—^ — and lo^ + is '
Process.
2a;2 + 3a; 4a:2_6a:_a;(2a; + 3) 2x(2x 3)
4x8 >< 12a;+ 18~ 4x^ ^ 6 (2 a; + 3)
_ x(2a: + 3) X 2 a: (2 a; 3) _ 2a:3
~ 4 a;8 X 6 (2 a: + 3) ~ 12 x *
Exercise 69.
Simplify :
^ a2 j2 c2 3a3 2h^ 7c^ ^ ^
^ Fc^'^c^ aV 4.c^ "^ 21 a^"" 5 ah' ^ r'
Sah^ 3«c2 Sad^ Sc^x^ 2() (^ x
2.
4crf 2hd "^ 9hc ' 5a^y^ 9a^y^
x+1 x + 2 x1 Za^x 10^
^' ^^^=~l^ x^1^ {x + 2f' 5 ^2a;24ic
a^ + 3 .r 4 2 x^^'Jx\\2 ^ m^  n^ m^n
^ _^ 9 a; + 20 ^ :i2 __ 5 ^ + 6 ' ^3 _ ^2^ ^3^ ^£
X'
6_
ALGEBRAIC FRACTIONS. 183
2/6 ^■\'f x + y
X
^ a^ + 2 3^ ij^ + 1/ s^  xy { 1/ "" a^  f
am fm (i\ m^ 4 w* ^i f _!!!: ^ \
* oTw \a~7wy' m^ + n^ \m—n m + nj
m^\mn nfi — n^ a^—(a^b)x{ab x^—<?
"' m^'^z ^mn{m\ny a^{a + c)x+ ac^ x^l^'
m° —
m^ + n^ f.. ^ ^
^' m^ — m n + n^ m^ + mn + n^ \ m — nj
9. g _ ? + 1 V^2 + ^ + 1) • Suggestion.
[e)i][(s)g(s)"0'
^^ fx a y h\ (x a V h\
10. ( + ?Ix( i + ]'
\a X b yj \a X b yj
\bc ac ab a J \ a + b + cj
a^ •\ ab — ac (a + c)^ — V^ ab — b^ — be
a:22a;" 63 ^ a^ + 3a:"40 ^ ?M^4^^+3 '
r a^> y'"* 1>.r (^'^?/^"')^ 1
8L Example 1. Find the quotient of . divided by ^'
fl c
Solution. Let x represent the quotient. Then t ^ 5 = «.
Since the quotient multiplied by the divisor gives the dividend,
184 ELEMENTS OF ALGEBRA.
we have x X . = j^. Multiplying both members of the equation
d c d a d ad
by  , we have a:X;X = TX, oraj^yX*
•^c' d c c^ he
Therefore, ^^^=:^X=^. Hence, m general,
To Divide a Fraction by a Fraction, invert the divisor, and
proceed as in multiplication.
Notes: 1. Since an entire or mixed expression may be written in fractional
form, the above method is api)licable to all cases. Thus,
_^a _ c _^a _ c ^__^c a _ a c _a 1 a
^ ' b"! ' l~l^a~ ~^'' b'^^~l^\~b c~bc'
2. It is usually better to change mixed expressions to fractional form before
performing the division. Thus,
(_a6\^/ ab \ _ <fi ^ b^ _ a'^ a + b __a^
" T+b) ■ V aT6/ ~ a + 6 ' aTl> ~ ^TTb ^ ~W~ ~ P '
^. ., ar214a:15, x^l2x45
Examples. Dmde ^,_^^^^^ by ^^^_^.
Process.
a:g14ar15 . g;212a;45 (x  15) (x + 1) , (x  15) (a: + 3)
a;24a;45 ' a;26a:27 ~ (x9)(x + 5) ~ (ar9)(a; + 3)
 (a^15)(a:+l) (re 9)
~ (a:  9) (a: + 5) ^ (x  15)
_ (a^15)Ca;+l)(a:9) x+ 1
" (a:  9) (a: + 5) (a;  15) ~ a; + 5 '
^. ., a?^ 1 , a: 1 1
Example 3. Divide ^+ibyp+
Process.
\y^ x) ~ \y^~ y x) ~ xy^ ' x y^
_ {x \ y) (x^  X y + y^) xy^
xy^ x^— xy + y^
_x + y __x
ALGEBRAIC FRACTIONS. 186
Exercise 70.
Divide :
2 a^ 2:1 y , a x^ y"^ Z m , 2 m
6 (g 6  ^/2) 2 62 2;3_y3 (a;  y)8
•^ a (a +6)2 '^^aCa^feZ)' a^ ^ f ^ {x + y^'
x^f xy ^ a^^xy+ip' ^  1?'
m8+8^m + 2' 2? ^ f ^ 3?xy\y^'
^ 2^:2+13^^15 22:2_^ 11^ + 5
5 ^^_Q by
4a:2_9 ^ 43^2
a:2 4 a: y + ?y2 x + y m
m
6. ;;o ^^H^ by — ^^; ^ ^ by  + .
x^ — xy)r]rx — y n^ m^ 71 m
yj X^y xy x+if xy X y
' X ^ y X + y ^ X — y' x \ y ^ y x'
x^^ (a + c)x + ac x^  o?
^ a?» + (6 + ^) a: + 6 c ^ ar»  62 •
a2 4.^>2_^2ft6~c2 « + 6 + c
^ c2a2_i,2+2a6 ^ 6 + ca'
10. a;8^ by a:; a262c2+2&c by ^^44^^
a:^ ^ a: ^ a + 6+c
11 ^6 . ^6 by 270; e— r by
7i6 + a:« ' 7i2 + a:2» fl^6__i ^ a^^a^^al
^^ x~^ — x~^. x~i { x~^
12. 0^8 by
2a:8 4a^2(xfa:*)
186 ELEMENTS OF ALGEBRA.
Exercise 71.
Perform the operations indicated in the following and reduce the
results to their simplest forms :
7 ic + 6 . a:2 + 6a; \ . . a;2 + lOa; + 24
48*
/x^7x + 6^ x'^ + 6x\ x^ + 10a; +
^* \x^h 3a; 4 ~ x^  8 x^J ^ a;^ 14a; +
/ x2  7 a; + 6 a;2 + 6a;\ a;'^ + 10 a; + 24
Process. (^^2 + 3 a; _ 4 "^ a;38a;2j >^ a;^ 14a; + 48
i [ (^  6) (a;  1) ^ a;(a; + 6) 1 (a; + 4) (a; + 6)
~ [(a; + 4) (a;  1) "^ x'^ {x  8) J ^ (a;  6) (x  8)
_ f a; 6 a;(x8) 1 (a; + 4) (a: + 6)
~ [a; + 4 ^ a; + 6 J ^ (a;  6) (x  8)
_ a; (x  6) (a;  8) (x + 4) (a; f 6) _
 (x + 4) (X + 6) (X  6) (x  8) ~ '^'
al a + 1 «2_i
X :
a + i a — l'a +
1' Va&~ a + &J * a262
\x + y X — y x^ — y^J \x \ y or — y^J
^2 _ ^ _ 20 x^x2 x^S6 ^ + 1
4. — 5 :t?— X 07r. o X
x^2d x^ + 2xS^x^6x ' x^ '\ 5x
?/4 a + h x^—3xy + 2y^ ^ {^ — vf
X 7 ; ZTt X
' a^h + ah {x \ y)^ x^ + y'^ ' ah
/af Z) « — &\ /« + & a  &\
max a^ — x^ h c } h x c — x ^ mx
noy c^ — x^ a^ + a X a — x ny
ALGEBRAIC FRACTIONS. 187
8 1 • PM I Mx ^"^ 1.
' x^y \_2\x + y x — y) x^y + xy^J
x^\x2 aP^5x+4: . f x^+Sx{2 x+3 \
^^' \6x62 * x^^l^J'^ax+a^' 8a^y^^ 21b"'^^y'^^
X 6^mH^a^ (a:  2)^ , ar^  4
^a' ,^24 ^ 8 ?w?i + a * (ir + 2)a*
12. ., o a X — =— i X '^
14. (a:* — jjHfa; J, by inspection.
15 (p  2 + ^2)  (?  I)' ^y inspection.
16. ic^— g — sfa; j \^ix ), by inspection.
6flg^>g , r 3a(m7i) . 5 4(ca;) ^ c^  a^ ^
m+n * I7{c + x) ' I 21afe2 * ^(m^n^ij
188 ELEMENTS OF ALGEBRA.
82. A Complex Fraction is one having a fraction in its
numerator or denominator, or both ; as,
n' , m *
n + —
n
Example 1. Eeduce  to its simplest form.
c
d
Solution. A complex fraction may be regarded as representing
the quotient of the numerator divided by the denominator. Hence,
a
h a ^ c a d _ a d
cj^l^d'^'b^'i^'bc'
d
h
a — 
Example 2. Reduce f to its simplest form.
m
Solution. Since the divisor is m, we have
6
\c — b m ac — h 1 ac — h
= [a — ~\ ^ m = 
x =
I c m cm
I
I m , m
Example 3. Reduce ~ » T » and — to their simplest forms.
^ 1 m n n
Process. — =zif = i x — = —
m n mm
m 1 n
Y = wir = m X ^ = mn.
n
1
m 1 1 1 n n
r=~"^~ = ~ X7 = . Hence, m general.
I m n m I m ' &
ALGEBRAIC FRACTIONS. 189
To Simplify a Complex Fraction. Divide the numerator by
the denominator.
Example 4. SimpUfy "*' " ^^' ~ '"' + '^^
m + n m — n
m — n m + n
Process.
m j n m — n ~ (m h ?i)2 — (771 — n)^ ~ 4 wm
m — n m 4 n (m — n) (m + n) (m — n) (m + n)
4 m^n* (m — n) (m + n) in n
X
(m«  n2) (m^ + n^) '^ 4 mn m* + n^
Example 5. Simplify ^ ^ iiJ2
xy x+y
Solution. Multiplying both terms by (x — y) (x + y), the L.C.D.
of their denominators, we have
2x1/
Notes : 1 . In many examples it is advisable to multiply both terms of the
fraction by the L. C. D. of its denominators at once.
2. If the terms of the complex fraction are complicated, the beginner is
advised to simplify each separately.
mp
Example 6. Simplify ^'+(^+^03:+ mn x^ + {m+p)x + mp
X* + (n + /?) X + n p
190 ELEMENTS OF ALGEBRA.
Process.
mp mn mp
v'^+(m^n)x + m7i x^h{m+p)x + mp _ {x\rm){x + n) (x+m)(x+p)
n — p ~~ n ~ p
x'^+ {n+p)x + np {x + n){x+p)
m n {x\p) — mp (x+n)
_ {x\7ri) {x+n) {x+p) _ inx (n ~ p) {x + n) (x+p)
n — p ~ {x + m) {x+n) {x+p) n—p
(x + n) {x+p)
rax
X + m
Example 7. Simplify T+~x'
1 '
lx + x'^ +
a:2_ 1
1 + X
X
Solution. Begin with the complex fraction x^—\ ' '^^"®»
a:2 _ 1 x+l , X x'^
i + x—— = ^, and ^5^n[ = rri ^'"^"^^^^y
i + ^^r
l +x^ (l + x^){l + x) ,^ l+x^
• ^= ,:8 + ^2+i ,and 1 x^
lX + X^+ ——r lX + X^ +
x^ + x^+l
Therefore
' 1 +x^ X
1 X x^ + x^+l
1  X + X^ + o 7
1 +x ——
X
=  (X8 + X2 + 1).
Notes : 3. A fraction of the form in Example 7 is called a Continued
Fraction.
4. To simplify a continued fraction, the student should always begin with
the last complex fraction in the denominator.
ALGEBRAIC FRACTIONS. 191
Exercise 72.
Keduce to their simplest forms :
a; + 6 + ^ 1 +  a: +  xy
^ j: — 6 wt c m n
1 ' 6 ^ ' mm n ^
X6 + 1 a;+ + 271
X + o m n X
, . 6^ 1 . 1 ni b m + n
a\b + ^  + — .
^ a n m n m 4:mn
n mm n S m^n
a;+la;l ar817a;+72
2 x1 ./J + 1 , 2:2 ^ 22 a; 4 120
^^ + 1 xV x"  21 a: +708
x^^ iiTTl a?» + 18 a; + 80
1 + « 4a 6 \a^; aj\x a)
l + a6 2a6 a; + a
J, + J_.JL
 mn mp np f x , 1— A_/ x \—x\
' m^(/i + pF ^ Vh^ ~^y \i+^"""^/
6.
m 71
1 1 1
^ 1 ' 1 ' ^1 •
«.' + J 1 X —
x +  1 + i ^ + — ^
a; a; a; — 1
192
7.
ELEMENTS OF ALGEBRA.
1 X2
1 +
l + a; +
'1^
1x
x2
x1
x +
y
X
X 1 , 1' _L
y^ y X
x — 2
2 xy
~ (^ + yf
1+
x+ 1
d +
m
1 +
^,y
(^'  yf X
ax
a? «2
X'
X
X
X x^
ir
+ 0 0++1
a a?
2/^ ^
X — y x" y^
y ^
10.
11.
w? + n^ 2 m
m^ — ii^ ' 7?< + n
~m n — m^ ^ m {• n
(m — iif' ' 'IV — n_
m — n
71 +
m — n
1 + m n
{in — n) n
1 \ mn
m
m
— n
1 
• m n
1 
(m —
n)m
1 — mn ^
fm n\
\n m)
83. Example. Find the third power of t;
Solution. Since an exponent shows how many times an expres
sion is taken as a factor, we have
'aJ^\
{aJ^y
J^nj  6" >^ &« >< ftn  (hny ~ h^n '
Hence,
To Find any Power of a Fraction. Raise both terms of the
fraction to the recjuirecl power.
ALGEBRAIC FRACTIONS.
193
Exercise 73.
Expand, by inspection, the following:
^(•^)' m' [G)'^a)T
/ 2aHixi y' f ix + ;/f y r m (x  y) !^
r (r+,,)(x,,) y r («i)»(«5)n « /a^a5)i\w
*• L "« + » J' L (2«+3t)i J' ^ ^ J
«C4?)U©©T=
+ 1
arff X
wi
ai
Jx
7«^
84. Example 1. Find the rth root of
6"
Solution. Since the rth power is found by takinp^ the numerator
and denominator r times as a factor, the rth root is found by taking
the rth root of each of its terms. The operation is indicated by
dividing the exponent of each term by r. Thus,
13
194
ELEMENTS OF ALGEBRA.
5«4
Illustration, y 243^ =^ ssTs^^isTs  3p • ^®^c®»
To Find any Root of a Fraction. Take the required root of
each of its terms.
Example 2. Find the square root of^2"2 — aa;+j + — + ^
Arranging according to powers of a, we have
X a
Process.
First term of the root squared,
First remainder,
First trial divisor, a^
a
First complete divisor, « + 
 times first complete divisor,
Second remainder,
Second trial divisor,
4 + ^ + ^^"^^ + ^^
x^ a^
a2 +
2a
X
2 a X
Second complete divisor, a'^ + — — 
—  times second complete divisor,
a8
+
a2
X^
X
rc2
ax
2 +
a2
aa;2 +
a2
Note. If we take —  for the square root of ^ , we shall arrive at the
result — p: f  .
2 a; a
ALGEBRAIC FRACTIONS. 195
Exercise 74.
Find the values of the following expressions :
1 Vy^286' V a^ ^ V 343 ar^* ^ ^243 ^.25" j •
f ni^n^ \^ ( Z2a^\\ ( (SA.m^n^x^ \\
\ a^^ ) ' \ b'^ ) ' \ 12oaH^7/y '
Find the square roots of :
^, .. ^^ . 4 . ^^^ . ^
Miscellaneous Exercise 75.
Reduce to lowest terms :
h(bax) + a(a + bx) 2^9 a^ + 7ix^ + 9 xS
(baxfh{a + bxf' a^ + Ta:^ 9 ^  7a;+ 8 *
21 x^ y^  S5 i/^z 12 x^z + 20 xyz^
l8x^z^21x^yS0y2^+S5xfz'
40r ^y*?>2 2^ yz^ 5i/^z^ + 4xs^
40:^28 36a:*22'5/2 + 4oa^2/8*
196 ELEMENTS OF ALGEBRA.
^3~6a;237;r + 210 ^ a;4»+10 ^""^ 35 rg2"+ 50 a;^+ 24
■ 0^+4x^^7 X2W ic3»+9^2n_.26a:" + 24
Find the values of :
_^2a;2/^2;^, .  ^a — ic
0. o x^ ] o when x = 4, v = k, z = 1: r:
when 2; =
a + b
„ x^ + y''^ — z^ + 2xy . ^ .
^ 2; — a ,x' — 6 , ft2 X X
7. — ^^ when x = 7;  +
b a a—b a b—a a+b
a^ (b — a)
when X = , ' , — ^ 
b (b + a)
, ^ ... ^ — 2 a + b a + b
8 < ^  + a2b ^^^'^ ^ ^ ~2~ •
(X — «V x
X — b ) X
9. i7^[_r + V ^ 2 J when x=^, ?/ = 1.
2ab\2bc^2ccl^2ad ' y(a_c)(J+c)+6(c&)(a2&)
when « = 3, 5 = 1, c =  2, t? = 6.
g^ + g c 4 &2 ^ {4^ai^yij^ ^^\ c
^2 — « c + 52 V'4a& — 522^ ~ a + 6 + cHfZ
when a = 4, & = 3, c = 1, c? = 7.
Divide :
12. !»J^by™x;^+gby?+2^.
ALGEBRAIC FRACTIONS. 197
1 , \ t:^ m^ . X m
13. m* i by i?i ; 4 r by  H
14 a^ + ^byx + ; a;3 ^ ^ + ^ _ _ by ^   .
15. a2_j2_,2_26cby ^44^'
16
• ^+iK^^^0''(^'^^)'^^'^
Factor :
Simplify :
20. 3x{y + [2..(y.)]} + i + ^^.
a;— 1 x—\ a; + 3 a: + 3
"3"*^^^ . 7 ~^T4
^■^ a; + 2 a; 4 2 * a;— 2 x—2'
'~T'^x^ 3 '^0^1
a\ l\ c\
3a5c a h c
' be h ac — ab la.1 1
a 6 c
198 ELEMENTS OF ALGEBRA.
23 ^ , ^ I ^^ + ^ ,
9^>2(4c2 a)2 Uc^{2auy^ 4.a^{U4.cf
' (2a+3&)216c2 + (36 + 4c)24ti2 + (2a+4c)'^9&2
\in/' j\m—n J \n^ J \m^\m7i+n^ J
h
a +
1 X { a 1 X — a i_L^
27. ^IZ+f! + 5^Z±Z; ^ X («^5«).
1 a + a; 1 a — x' h ^ '
a 0^ + Q(^ ^ <x2 _}_ ^2 ^
h
a — h — c h — c — a c — a — h
a^—ac — ab + bc b'^—ab — cb+ac c^—bc—ac+ab
X X
11 1
30.
2/ + r
ALGEBRAIC FRACTIONS. 199
31.
/ 3 2: + 3:3 X2
yif 32:V
9^ _ 33  ar^ 3
^ 32:2+1 ahc
a:3_3a:'^ 2:^ (a^a:)^ hc^ ac ab
Sab + c 62 a162 a^b^ (a^b^y
a ^b — c * a^ ab~^
aH~^ / a2 6^ Y
'^2:'"+?/" '^2^^'"+?/2'* aU^ &Ki ai(^
2:+y" x^+y^ jici
10 2; 7/ 3^2+ 10a:3y _^ lQ2:3y _3_
15i/^+102;//2 + :30y+202:y ' 452/ + 302:y ■*" y+2 '
 (f)(i^r')H"')(Sa?)
35 ^ + ^'' ^ + ?^^\.f + y') ■ \ff x)
(x + y)^ — xy
(x — yf\xy
r g<?/ ^ rt2+a?/ 1 To^oV g^2qg.v+fl27/n
37. * 1
Sax — 5b y ax '^by'^
1 ^""3a2;2fey
38. i^+l+ 1
1 + i— 1 +T— 1 +
t z X \ z X {■ y
200 ELEMENTS OF ALGEBRA.
39. (a + 2> + .)^ + ^ + j ^^^
(c  a) {ah)'^ {a h) {h  c)'^ {h  c) (c  a)'
fa^—ij^ lOx^—lSxy—Sy^ 2a:^ + xy^+xy+y^\
\x^—jf l<)oiP'—Zxy—y^ 2x^ + xy—%f' )
xy — y'^— 2x+2y
~ 2a;?/
(a + hf  63 {h + cf  g g (ci + c)3  63
44.
45.
(a + i) — c h \ c — a a ■\ c — h
11 1
a(a — &)(a — <?) b(b — a)(b — c) abc
2a + n a + b + 71 m + n—a
am+ab—bm—o? ab+bm—am—b^ m^—bm—am+ah
46. 7 rc^ X + TT, TT^ X +
a(a—b)(a — c) b(b — a){b — c) c{c—a)(c—b)
Queries. Why does changing the sign of one factor of either term
of a fraction change the sign of that term ? Will it change the sign
of the fraction ? Why 1 When the denominators have no common
factors why multiply both terms by the product of the denominators
of all the other fractions 1 Why does the process of reducing to
forms having a common denominator not change the value of a frac
tion ? How prove the methods for addition, subtraction, multiplica
tion, and division of fractions 1
FRACTIONAL EQUATIONS. 201
CHAPTER XV.
FRACTIONAL EQUATIONS.
85. Example 1. Solve j^  147^315 = "gT 30~
J_
■*"105*
Solatioii. Multiplying each member by 210 (the L. C. M. of
15, 21, 30, and 105), transposing and uniting like terms, we have
— j = 5 + 30 X. Multiplying each member of this equa
tion by X— 1, transposing and uniting like terms, we have 25a; = 100.
.. x = 4.
Proof. Substituting 4 for x in the given equation, we have
65X4 72X4" _ 1+3X4 _ 10 X 4  11 1
15 "14(41) ~ 21 30 "^105*
or, — ^^ = — ^yV' which is an identity. Hence,
To Clear an Equation of Fractions. Multiply each member
by the L. C. M. of the denominators.
o c, 1 2x+U 2fxl xl
Example 2. Solve =— ^  ^ = ^ •
5 50 z — 10 2^
Proceas. Multiply by 5, 2 x + 1^  loT^ = 2 x  1.
Transpose and unite,  r^ — — ^ = — 2^.
Clear of fractions,  (2^ a:  1) =  25 x + 5.
Transpose and unite, 22.6 x = 4. .*. x = ^W*
202 ELEMENTS OF ALGEBRA.
Note. In solving a fractional equation, where some of the denominators
are simple and some are compound expressions, it is better to multiply each
member of the equation by an expression which will remove the simple denom
inators tirst, then transpose (if necessary) and unite like terms. Similarly
remove the compound denominators of the resulting equation.
Exercise 76.
Solve the following equations :
■ X ^ Vlx~ 24' 2 '6x ~ ^ ~ 'Ix '
6a;+13 _ 32:45 2x 2^5 x?> _4a:3 ^
15 5x25 ~ 5 ' 5 "^ 2i^=^ ~ 10 i^
9 2^+5 8^7_36^+15 lOJ
14 ^6^+2" 56 "^l4"
4a:+3 73;29 _ Rrr+19 3,x+2 _ 2a:l ^ a;
9 "^5:^12" 18 . ' 6 3a:7 2*
^ \^xTl ^ I+I62; ,. 101642;
^ ^9^^6^+2^ + ^4^^^^— 24
6.
18 2^+10 72 a; + 30 _ 20.5 16 2;  14
42 168 ~ 42 18aj+ 6
1 2 _2;+2 4(2;43) _ 82;+37 72;29
^ 2"^'2:+2~ 22; ' 9 "" 18 52; 12
^ 2 2: + 8r} _ 13 2;  2 , ^ _ 7^ _ 3^+16 ^
9 17 2^32 ' 3~ 12 36
g+ 1 22:4 22; 1 x2 x4:
T5~ 72;16~ 5 ' .05 0625
FRACTIONAL EQUATIONS. 203
86. Frequently it is better to unite some of the terms before
clearing the equation of fractions. Thus,
X
^^ ~ 3 16x4 4.2 23
Example 1. Solve ^^y + g^.^^ = ^ + ^Tl *
X
^^~3 23 16X+4.2 ^
Process. Transpose, — ^  ^q[ + 3^+2" = ^
,, . ,, ^~3 I6X+4.2 ^
Unite like terms, jTTj" H — g , « = 5.
Free from fractions, 4+Vxa:^+16x2+20.2x+4.2 = 15a:^25a:+10.
1.6x
Transpose and unite, — g— = 1.8.
.. a; = 3f.
Example 2. Solve ^—^  jqjg " ^a^Ti = 0
Process. Multiply by x^  4, (x + 2)  (x  2)  (x + 1) = 0.
Simplify, x+3 = 0. .. x = 3.
Notes : 1. If a fraction is preceded by the — sign, in clearing the eqiiation of
fractions, care must be taken to change the sign of each term of the numerator.
In such case it is convenient to enclose the numerator in parentheses before
clearing the equation of fractions.
2. The student should be careful to observe that he can make but two
classes of changes upon an equation without destroying the equality :
I. Such as do not affect the value of the members.
II. Such as affect both members equally.
Thus, in the above process, the first operation affects both members equally;
and the second, that of uniting like terms, does not affect the value of the
members.
4 2 5 24
Example 3. Solve ^j^  ^^^^ = ^^  ^^ .
Solution. Transposing, ^  g^ = ^  2F+2 '
204 ELEMENTS OF ALGEBRA.
Simplifying each member separately, we have
3 _ 11 1
2 (x + 3) 2 (a: + 1) ' "^ a; + 3  2 (a: + 1)
Clearing of fractions, we have 2 (a; + 1) = x + 3. r. x=l,
c^ A oi ^ — 4 x — 5 x — 7 x — 8
Example 4. Solve r
x5 xQ xS x9
Solution. Reduce the fractions to mixed expressions,
1 1 1 1 ^ , . ,
or ^3^  ^^^^ = ^jg  ^^^ • Reducmg the terms in each
member separately to common denominators and adding, we get
 (a:5)(a;6) =" ~ (a:  8) (a;  9) ' ^^^^""^ *^^« ^^"^^^^^ «f
fractions, we have —{x — 8) (^ — 9) = — (x — 6) (x — 6). Simplify
ing, transposing, and uniting like terms, — 6 a: = — 42. .'. x = 7.
(2 a; + 3) a; J^
2a;+ 1 "^ 3a;
dx A 3^ X
Process. Reduce ^^ , i to a mixed expression,
Transpose and unite,  ^j:^ ==  3^ '
Clear of fractions, — 3a; = — 2a:
Therefore, a; =. 1.
n. r. , 5a;64 2a:ll 4a;55
Example 6. Solve
a: 13 xQ x14: x1
Process. Reduce the fractions to mixed expressions,
FRACTIONAL EQUATIONS. 205
Simplify each member separately,
7
(x  13) (x  6) ~ (2:  14) (x  7)
Divide by 7 and clear of fractions,
x2  21 a; + 98 = a?2  19x + 78.
Therefore, x = 10.
Exercise 77,
Solve the following equations :
12 1 29 x + 4 x+6
a; ' 12a; 2.4* 32;8 3a:7
3a:+l a;2 6a:+l 2 a;  4 2a;l
3(a:2) a;l' 15 7a:16 5
x\25 ^ 2x + 75 5 4 _ 3
a;5 "" 2a:15' 1  5 a: "*" 2 a;  1 "" 3a:  1
6a;+8 2a;+38 _ x^^x+1 a^+x{l _ ^
2a:+l a;+12 ~ * a;l "^ a: + 1 " ^^'
^7_2^15 1 J 2 1__
a:+7 2x'6"*"2a;+14"~ '1a: 1+a: lx^~^'
3 30 3.5
4  2 X 8 (1  a;) 2  ./; 2  2 a;
6^7i l + 16a: , 12180.
^' 1312a:+'^^+ 24 ^ ^^^ 3
a;— 1 a; — 5 a; — 4a;2
aj — 2 a; — 6 a;— 5 a* — 3
5a;8 6a;r44 10a;8a;~8
' a;2 ar — 7 a; — 1 "" a; — 6 '
206 ELEMENTS OF ALGEBRA.
xl x+1 _ 2(x^ + 4xtl)
^^ • x2'^ x + 2~ {x+2f
, , 30 + 6 rr 60 + 8 a^ ^ . 48
11. ; — H ; — rt — = 14 +
X + I X + O X + 1
.6a:+.044 .5^.178 _ .3a^l _ .5 + 1.2 a;
■^^' A .6 ^^^^ .5xA~ 2x^1
2xZ Ax .6 1lAx _ .7{xl)
13.
.3ic.4 .06;:c.07' x + .2 .1  .b x
87. A Literal Equation is one in which some known
number is represented by a letter; as.
X X
Example 1. Solve — f
m n — m m jr n
Process. Clear of fractions, x {n"^ —m^)+x (m^ +mn) = m^{n m)
Unite like terms, {n^ +mn)x = m^{n — m).
m^(n—m)
Divide hy n(n + m), x = ^^^^^_^^^y
Example 2. Solve (xm) (xn) — (xn) {xa} = 2(xm) (ma).
Process. Simplify, transpose, and unite,
Sax — 3mx=: — 2m^+ 2am ~ mn \ an.
Factor, 3 (a m)x = (a — ?«) (2m + n).
2m + n
Divide by 3 (a — m), x = ^ — •
a23&x ,„ , 6 6a:5rt2
Examples. Solve ax ab^ = ox\ ^
a 2a
bx + 4a
4
Process. Clear of fractions, simplify, transpose, and unite,
4a^x3abx = 4a^b^ 10 a^.
Factor, a (4 a  3b) x = 2 a^ (2b^  5).
Divide by a (4 a  3 6), • x= ^\_.^f^ '
FRACTIONAL EQUATIONS. 207
_, ax ~b bx — a a — b
Example 4. Solve , . — l^ , ,. = /„ ^ , h\ /k ^ ^ „\ '
ax {■ ox + a {ax + o) {ox + a)
Solution. Reducing the terms of the first member to mixed ex
/ 2b \ f 2a \ ab
pressions, we have [l  ^^ j  [l ^^^j = (^^^^^^^^^^^ •
Uniting like terms and reducing the fractions to a common
denominator, adding and factoring their numerators, we have
2(a{b) {ab)x ab ^,, . ^ , .
7 . iv /I. — ; — 7 = 7 , .V .. — ; — ; . Clearing of fractions,
{ax + b) {bx + a) {ax + b) {bx + a) " *
2{a + b) {a — b) X = a — b. Therefore, x — ^ . . v •
Notes: 1. Example 4 may be solved by clearing tlie equation of fractions.
The solution is presented as an expeditious method.
2. If the student cannot readily discover a special artifice, be should clear
the equation of fractions at once.
3. Known terms are called absolute terms. Thus, in the equation mx^
•\ nz \ a = 0, a is called the absolute term.
a i b a b
Example 5. Solve ; = 0.
x — c X — a X — b
Process. Clear of fractions,
{a\b){xa){xb)a{xb){xc)b{xa)(xc) = 0.
Simplify, transpose, and factor,
x{ac + bc  a^b^) = ab{2c ab).
Tx. ., , , « .„ ab{2cab)
Divide by a c + 6 c  a*  62, x = 7^; 5 — A*
^ ' ac + bc — a^—b^
_fl &(a + 62c)
°' ^~a2 + 63c(a + 6)'
Exercise 78.
Solve the following equations :
10. e a 6 1
X a ^x
2. \0hmx — ^an = 2am — hhnx\ = r*
a X X
208 ELEMENTS OF ALGEBRA.
^ 7n? n 4:71^ m a h „ ,«
X A X 4: ox ax
4..(.«)(^±^J=^(.).
5. ^^  ^^ +2 = 0; (xa)(xh) = {x'ahf.
^ a{b^x + a^) ax^ 2fx A Zfx \
6. ^— V ~ aca; + ,;  + 1)= — 1).
hx b S\a ) 4\a /
3 ah — x^ 4:X — a c x^ — a a — x 2 x a
^ ^ ^ ^ ^^ w^.^ l^ ^ ^
' c hx ex ' hx h
h
^ X — m 0? — mx — V? ^ n?
L — 1 •
VI mx — n^ mx — n^
Miscellaneous Exercise 79.
Solve the equations :
X 07+1 ^ — 2x'^ac hc_ ,
9 ^ ~1> 1  9aj' h~x~'^~^ '^ '
ax+h Sh ^ a^x^ + h^
ax — h a X + h~ a^x^ — b^'
X X ■\ \ _x — ^ X — ^
' x — 2 X — 1 X — ^ X — 1 '
2(2a;+3) 6 bx+\
4 1
639^ 1x 2% Ax
FRACTIONAL EQUATIONS. 209
5 1 I 1 1 0.
a (6 — a;) b{c — x) a(c ^ x)
6. (2a:^)rar+^^ =4a:^^a:Vj(a4a;)(2a + 3a:).
17 __ . _ 105 +10a; _ .^
^* a; H 3 ^ ~ 3 a: + 9
8. (.^3)^_^^^) = 7.(3.?i^)).
a?— a a + a; 2aa;_ 1 ^ 1 __ a — h
' a—b a {• b a^—l^~ * x — a a;— 6 x^—ab
10. 3— + — j — = 2 a;; =c(a — ^)) + .
a; — la; 41 x ^ ^ x
_ a:+ 2 , a: 7 a: + 3 x  ^
11. h ■= — — — z = 7 •
a; X — o X + 1 X — 4:
135 a;  .225 .36 .09 x  .18
12. .15 a; +
13.
.6 ~ .2 .9
x—a x—a—1 x^b x—b—1
a;a — 1 a? — a — 2 x—b — l x—b — 2
^ . SO a — bx 9 n — ax 6 m — nx
14. = 5 ^ = 0.
,^ 4m(a25./2) ^ 5 m (J^  2 a;)
8a: 4
X — np X — mp X — mn
16. p = p.
^ m n V
14
210 ELEMENTS OF ALGEBRA.
3 & (a; — a) a; — &2 ^ & (4 a + c a?)
5 a 15 6 ~ 6a
mx — n mx \ n
c^ — Sdx(P+2cx X X
' c^+odx d^—2cx~~* m ~~ n
n ',n
m x 7n(x—m) x(x + m) mx
x m x{x + m) m{x—m) m^—x^
Queries. Upon what principle is an equation cleared of fractions ?
How is it done '? Why change the signs of the terms of the numera
tor of a fraction, preceded by a minus sign, when clearing of fractions '?
Upon what principle (give four different explanations) may the signs
of all the terms of an equation be changed ?
Exercise 80.
1. The second digit of a number exceeds the first by 3 ;
and if the number, increased by 36, be divided by the
sum of its digits, the quotient is 10. Find the number.
Solution. Let x — the digit in tens' place.
Then a; + 3 = the digit in units' place,
and 2 a; + 3 = the sum of the digits.
Therefore, 10 a: + a: + 3, or 11 a: + 3 = the number.
lla:+3 + 36
Hence, — ^ — r5 — = 10. .. a:= 1. lla:+ 3 =14, the number.
PROBLEMS. 211
2. The first digit of a number is three times the second ;
and if the number, increased by 3, be divided by the differ
ence of the digits, the quotient is 17. Find the number.
3. The first digit of a number exceeds the second by 4 ;
and if the number be divided oy the sum of its digits, the
quotient is 7. Find the number.
4 The second digit of a number exceeds the first by 3 ;
and if the number, diminished by 9, be divided by the
sum of its digits, the quotient is 3. Find the number.
5. A can do a piece of work in 7 days, and B can do it
in 5 days. How long will it take A and B together to do
the work ?
Solution. Let x = the numler of days it will take A and B to
gether.
Then  = the part they do in one day ;
but = = the part A can do in one day,
and e = the part B can do in one day.
Therefore, = + ^ = the part A and B can do in one day.
7 o
Hence,  = ^ + ^. Therefore, x = 2\^.
6. A can do a piece of work in 2 J days, B in 3 days,
and C in 5 days. In what time will they do it. all work
ing together ?
7. A can do a piece of work in a days, B in 6 days,
C in c days. In what time will they do it, all working
together ?
212 ELEMENTS OF ALGEBRA.
8. A and B together can do a piece of work in 12 days,
A and C in 15 days, B and C in 20 days. In what time
can they do it, all working together ?
9. A and B together can do a piece of work in a days,
A and C in 6 days, B and C in c days. In what time can
they do it, all working together ? In what time can each
do it alone ?
10. A tank can be emptied by three pipes in 80 min
utes, 200 minutes, and 5 hours, respectively. In what
time will it be emptied if all three are running together ?
11. A sets out and travels at the rate of 9 miles in 5
hours. Six hours afterwards, B sets out from the same
place and travels in the same direction, at the rate of 11
miles in 6 hours. In how many hours will he overtake A ?
Solution. Let x — the number of hours B travels.
Then x + 6 = the numher of hours A travels;
also, I = the numher of miles per hour A travels,
and i^ = the number of miles per hour B travels.
Then, y^ x = the number of miles B travels,
and I (a: + 6) =r the number of miles A travels.
Hence, V" ^ = f (^ + 6). Therefore, x = 324.
12. A man walked to the top of a mountain at the rate
of 2 miles an hour, and down the same way at the rate of
3^ miles an hour, and is out 13 hours. How far is it to
the top of the mountain ?
13. A person has a hours at his disposal. How far
may he ride in a coach which travels b miles an hour, so
as to return home in time, if he can walk at the rate of c
miles an hour ?
PROBLEMS. 213
14. In going a certain distance, a train travelling 55
miles an hour takes 3 hours less than one travelling 45
miles an hour. Find the distance.
15. The distance between London and Edinburgh is
360 miles. One traveller starts from London and travels
at the rate of 5 miles an hour ; another starts at the same
time from Edinburgh, and travels at the rate of 7 miles an
hour. How far from London will they meet ?
16. The distance between A and B is 154 miles. One
traveller starts from A and travels at the rate of 3 miles
in 2 hours ; another starts at the same time from B, and
travels at the rate of 5 miles in 4 hours. How long and
how far did each travel before they met ?
17. The distance between A and B is a miles. One
traveller starts from A and travels at the rate ot 711 miles
in n hours ; another starts at the same time from B, and
travels at the rate of b miles in c hours. How long and
how far did each travel before they met?
1 8. If it takes m pieces of one kind of money to make
a dollar, and ?i pieces of another kind to make a dollar,
how many pieces of each kind will it take to make one
dollar containing c pieces ?
19. The denominator of a certain fraction exceeds the
numerator by 6 ; and if 8 be added to the denominator,
the value of the fraction is J. Find the fraction.
20. A can do a piece of work in 2 m days, B and A
gether in n days, and A and C in m + ^
time will they do it, all working together ?
together in n days, and A and C in m + ^ days. In what
214 ELEMENTS OF ALGEBRA.
21. In a composition of a certain number of pounds of
gunpowder the nitre was 10 pounds more than ^ of the
whole, the sulphur was 4^ pounds less than J of the whole,
and the charcoal 2 pounds less than ^ of the nitre. Find
the number of pounds in the gunpowder.
22. A broker invests  of a certain sum in 5 % bonds,
and the remainder in 6 bonds; his annual income is
$180. Find the amount in each kind of bond, and the
sum.
23. A broker invests — th of a certain sum in a % bonds,
n
and the remainder in c % bonds ; his annual income is b
dollars. Find the amount in each kind of bond, and the
sum invested.
24. At the same time that the westbound train going
at the rate of 33 miles an hour passed A, the eastbound
train going at the rate of 21 miles an hour passed B ; they
collided 18 miles beyond the midway station from A.
How far is A from B ?
25. A person setting out on a journey drove at the rate
of a miles an hour to the nearest railway station, distant h
miles from his home. On arriving at the station he found
that the train had left c hours before. At what rate should
he have driven in order to reach the station just in time
for the train ?
26. A merchant drew every year, upon the money he
had in business, the sum of a dollars for expenses. His
profits each year were the nth. part of what remained after
this deduction, but at the end 3 years he found his money
exhausted. How many dollars had he in the beginning ?
SIMULTANEOUS SIMPLE EQUATIONS. 215
CHAPTER XVL
SIMULTANEOUS SIMPLE EQUATIONS.
88. Simultaneous Equations are such as are satisfied by
the same values of the unknown numbers.
Thus, 3 X + y = 9 and 5a: — 2y = 4 are satisfied only hy x = 2
and y = S.
Elimination is the process of combining simultaneous
equations so as to cause one or more of the unknown
numbers to disappear.
This process enables us to fonn an equation containing but one
unknown number. The equation thus formed can be solved as
shown in the preceding chapter.
Hote. There are only three methods of elimination most commonly used.
Elimination by Addition or SnbtractioiL
89. Example 1. Solve the equations : 5 3a; 5?/ =13 (1)
^ l2x + 7y = S\ (2)
Hote 1. The abbreviations (1), (2), (3), etc., read "equation one," "equa
tion two," etc., are used for convenience to distinguish one equation from
another.
Solution. To eliminate x we must make its coefl&cients equal in
both equations. Multiplying the members of (1) by 2, and those
of (2) by 3, we have
5 6 X  10 y = 26 (3)
i6a; + 21y = 243 (4)
216 ELEMENTS OF ALGEBRA.
Subtracting the members of (3) from the correspojiding members
of (4), we have 31 y = 217. .'.y = 1. Substituting this value of y
in (1), we obtain 3 a;  35 = 13. .. x= 16.
VerifiGation. Substituting 16 for x, and 7 for ^ in (1) and (2),
we have 54835 = 13 (1),
we nave "[gg, 49^31 (2),
, , identities.
32 + 49 = 81 (2),
Votes : 2. In this sohition we eliminate x by subtraction. But suppose we
wish to eliminate y instead of x. Multiply (1) by 7, and (2) by 5, then add
the resulting equations, and we have 31 ic = 496. . •. ic = 16. This value of x
substituted in (1) gives y = 1.
3. When one of the unknown numbers has been found, we may use any one
of the equations to complete the solution, but it is more convenient to use the
one in which the number is less involved.
4. It is usually convenient to eliminate the unknown number which has the
smaller coefficients in the two equations. If the coefficients are prime to each
other, take each one as the multiplier of the other equation. If they are not
prime, find their L. C. M., divide their L. C. M. by the coefficient in each equa
tion, and the quotient will be the smallest multiplier for that equation.
Example 2. Solve the equations : 515^ + ^7^ = 92 (1)
^ ( 55 a:  33 7/ = 22 (2)
Solution. Multiplying the members of (1) by 11 (the quotient
of 165 divided by 15), and those of (2) by 3, we have
5 165 a; + 847 y = 1012 (3)
Xl^bx 99 2/= 66 (4)
Subtract the members of (4) fron the corresponding members of
(3), 9461/ = 946. .. ?/= 1. Substitute this value of y in (1),
15ar+77 = 92. .. a; = 1.
Proof. Substituting 1 for x, and 1 for ?/ in (1) and (2), we have
5 15 + 77 = 92 (1)
1 55  33 = 22 (2)
Hence, both equations are satisfied for a: = 1 and 2/ = !•
Example 3. Solve the equations : 5 ^7 a:  12 !/ = 289 (1)
^ ( 55 a; + 27 2/ = 491 (2)
SIMULTANEOUS SIMPLE EQUATIONS. 217
Process. Multiply (1) by 9, 693 x  108 y = 2601 (3)
Multiply (2) by 4, 220 x+lOSy= 1964 (4)
Add (3) aiid (4), 913 x = 4565. .. x = 5.
Substitute this value of x in (2), 275 + 27 y = 491. .♦. y = 8.
Prool Substitute 5 for x, and 8 for y in (1) and (2), and we
have J 279 = 279 (1), .^^^^^
\ 491 = 491 (2),
Let the student supply the method from the solutions.
Exercise 81.
Solve the following simultaueous simple equations
1.
3a; + 4y=10.
Ux"+ y= 9.
8.
(Jy + Ja; = 26.»
(fy + .; = 25.
2.
Sx y = 34.
a; + 8 y = 53.
9.
( .25 x + 4.5y = 10.
1. 75 y. 15 a; = .9.
3.
4
10 a: + 9y = 290.
12 2;lly = 130.
7 y  3 a; = 139.
2x + 5ij= 91.
10.
J 3^ 2 ''
l2 + 3 = ^
5.
{6x5y = 7.
\ 10 a; 4 3 7/ = 11.
11.
( .5 a: + 2y= 1.8.
1 .5 y  .8 a: = .08.
6.
9a;4y = 4.
15 aj + 8 y =  3.
12.
(7a:+^y = 99.
1 7 y 4 j a: = 51.
7.
9 y 42 a; =15.
4y47a;= 3.
13.
r Jaj4 3y = 22.
1 l\x\y=20.
♦ Clear of fractions first.
218 ELEMENTS OF ALGEBRA.
Elimination by Substitution.
90. Example. Solve the equations : HaJ + 32/ = 22 (1)
^ l5x7y= 6 (2)
Solution. From (2), x = — — ^ (3). Since the equations
o
are simultaneous, x means the same thing in both, the substitution
of this value of x in (1), will not destroy the equality. Hence,
4/ — F~^) +3^ = 22. Clearing of fractions, transposing, and
uniting like terms, 43 2^ = 86. .'. y = 2. Substitute this value of
y in (3), x = 4.
Let the student supply the method.
Exercise 82.
Solve by substitution :
1
2.
3.
x+Sij^U. ^ l^x + iy = 7.
7 x + 4:y = 29. {S7J + 4cX = SS.
Sx+ ?/ = ll. \5x+62j=61.
l^V^x = 21. l3+2 = l
I .08 2/  .21 a = .33. ( 3 y  4 a; = 1.
I .7z + .12y= 3.54 I 3 a;  2 ?/ = 1.
" ~ *• 10. I 11 ^
^ = 0.
^ = 1.
SIMULTANEOUS SIMPLE EQUATIONS. 219
Ml2/7^ = 37. (10a: = 9 + 7y.
^^ 8y + 9a: = 41. U2/ = 15a!7.
12. < aud verify.
Elimination by Comparison.
91. This method depends upon the following axiom :
6. Things equal to the same thing are equal to each
other.
Example. Solve the equation . )^^^y=^ 0)
^ (7x4y = 8^ (2)
Solution. From (1), x = i±A^ (3). From (2), x = ?i±l^.
Since these equations are simultaneous, x means the same thing
in both, — ^ = ^ ^ . Solving for y, we have y = 4. Sub
7 1 + 20
stituting this value in (3), x = — tj — = 3^.
Let the student supply the method.
Exercise 83.
Solve by comparison :
5a;+6y = — 8. j6x+l5y = — 6.
3x + 4y = 5.
(6x+ l5y = {
I 8 a;  21 y = 74
\}x{iy = S.
(^x + 3y = 51. rSy .7x =.4.
' \7x+2y = 3. I .02 y + .05 a: = .2$
12^7^=17.
• U2:+8y = 20. ^
220 ELEMENTS OF ALGEBRA.
(l.lx l.Sy =0.
^ t.l3:r .11 7/ = .48. \l +
0^ = 3 2/ 23. ^^1 5 2 ~'^*
12. { ^
1 = 42.
r.30^.772/=:2.95. h^ 4.y^4o
1 .20 a: +.21^=1.65. ^8"^ 9
92. Each of the equations should be reduced to its simplest form,
if necessary, before applying either method of elimination.
Notes : 1. An expeditious method, for the solution o^ particular examples,
is that of first adding the given equations, or subtracting one from the other.
2. Usually, in solving examples of two unknown numbers, it is expedient
to find the value of the second by substitution; but this is by no means
always so.
Example. Solve :
2y + 42:2tf 10i/5fx18
3a:+.y 13?/37^_ 99a;t/ 10a:+.25i/10.5
L~T2^+ 44 ^^+ 22~^~ 33 ^^^
Process. From (1), 127 y+ 59 x= 1928 (3)
From (2), 59 ^ + 127 rr = 1792 (4)
Adding (3) and (4), 186 y + 186 a; = 3720 (5)
Dividing (5) by 186, y+ a; = 20 (6)
Subtracting (4) from (3), 68?/ 68 a; = 126 (7)
Dividing (7) by 68, y  x = '2. (8)
Adding (6) and (8), %y =22. ..2^=11.
Subtracting (8) from (6), 2x~ 18. .. a:= 9.
Solve
SIMULTANEOUS SIMPLE EQUATIONS. 221
Exercise 84.
1 fy(^ + 7) = a:(y+ 1). (2y + Ax =1.2.
I 2y+20 = 32;+ 1. XsAy .02x= .01.
r(y+l)(2:+2)(y + 2)(2:+l) = l.
^ \ 3 (y + 3)  4 (2: + 4j =  8.
f .3 2: + .125y = a: G. rx4y = 3.
l3a:.5y = 28~.25y. I aj + v =32.
.5y = 28~.25y. {x + y
6. ^
'4:X + Sy 2 y \7x _ XS
To 24 ^"^ 5
9a;+52/ — 8 a; + y _ 7y 4 6
12
4
9. ^
<^ 5 + y 12 + a;
l2a;+ 53^ = 35.
10. ^
3y10(a:~l ) ^jy . . _ ^
6 ^ 4 "^ ^  "•
' 4a;3y7 _ 3_^ _ 2 .y _ 5
5 " 10 15 6 *
y~l , ? _ 3y _ 7/^^ 4. ? 4. JL
L 3 "^2 20 15 "^ 6 ■*■ 10
3 5 ~ 4 * U"^il 33
222 ELEMENTS OF ALGEBRA.
r:. + l(3^.yl)^i+f(2/l).
f2x _Sy — 2 _ _ 4: + X y — x
14. ^ 18
by
\2x.
2Zy X
2x^ =
a? + 43 X — y
2.4 X + .32 y
\ X ]r "i^o X — y
.36 X  .05
\'.
13.V
7
= 2i.
3^/ +
11
^9 =
= — ii;.
« ^ _L
2.6 +
.005 3/
17.
.5 ' .25
04 2/ + .1 .07 a? .1
18.
.6
_ Zx2\,y ^ IBy + jx
^ 11 "^33
2xh3y x5 11 :?/ + 152 3 a; + 1
r
?/  2 10  y ^  10
19 ^ ^ ^ ^
■^22; + 4 0^ + 4?/ +12
I
8
21 H(2^+ 72/) 1=1(20.62/+!).
I a? =: 4 2/.
SIMULTANEOUS SIMPLE EQUATIONS. 223
Suggestion. Multiply the members of the first equation by 2,
transpose, and unite like teims ; then clear the resulting equation of
fractions. Multiply the second equation by 3, transpose, and unite
like terms; etc.
23. <
24. <
25.
r2 , y ^ 3?/ 1
,11 + 2 = 12.. 6.
f 6x + 9 3x5 _ 3x + 4:
4 +4y~6~ i"^ 2
8a; 4 7 Sx 6y _ 94a ;
^10 2a:8~ 5
16 + 60x _ 16xy 107
3yl ~ 5 + 2y •
Suggestion. Multiply the members of the first equation by
5 + 2 y, transpose, and unite like terms ; then clear of fractions ; etc.
26.
27.
X — y __1
X \ y 5 *
13 3
y+2a;+3 4?/ — 5a;+6
3 ^ 19
6y5a; + 4~ 3y + 2a;+ l'
ryx = l. r5(y+3) = 3(ic~2) + 2.
28. I y+ 1 _ y1 _ 6 29. ^ 2 ^ 3
U— 1 X ~ 7' i^y+3~a: — 2*
224 ELEMENTS OF ALGEBRA.
30.
31.
32. <
i(2 2/ + 7^) 1 = 1(2 2/ 6^+1).
U =
y
6y224a^+130
2y4:x+ 3
151  16y _ 9 0^3/ 110
4^1 3a;4 '
^4a; + 22/ 4a;+53/
""16 31 "" •
2^+j/ 3 ?/  2 0? 36
~5 +""~6 =y
r5.T + 202/ = .l. 04 J 2/^ 3"
^'' \ll^ + 302/ = .9. l^+_^±i_7^Q
V2/ — a; — 1
35.
r 2a;— .5y 5jy— 19a;— 15 „ ^— a; + 2
93. Fractional simultaneous equations in which the unknown
numbers occur in the denominators as simple or like expressions^ are
readily solved without previously clearing of fractions. Thus,
Example 1. Solve:
^'h
21
y
=
10
20
6
X
y

2
(1)
(2)
Solution.
10 3
Dividing the members of (2) by 2, we have — —  = 1 (3). Mul
70 21
tiplying the members of (3) by 7, — — — = 7 (4). Adding the
X y
SIMULTANEOUS SIMPLE EQUATIONS. 225
members of (1) and the corresponding members of (4), we have
. X = 5. Substituting this
85 5
— = 17. Dividing by 17,  = 1
X X
value of X in (1), gives  = 1: .. y = 3.
Note If we cleared these equations of fractions they would give the pro
• liict xy, and thus become quite complex. In the solution of this particular
.lass of examples it is always easier to eliminate one of the xmknown numbers
without clearing of fractions.
Example 2. Solve:
2y^rx = ^
2 4 _
2 20 136
Process. Multiply (0 by i, 3T + g^ = ^
Subtract (3) from (2),
Simplify (4),
y
4_
5x
2,1 x
20
■ 27 X
8
135 X
208
9
208
9 •
2 2
Substitute in (2), ^  312 = 8, or « = 304.
^y ^y
(1)
(2)
(3)
(4)
^~ 390'
y =
456
Exercise 86.
Solve:
1. <
2.
r2 1 ,^
 +  = 10.
X y
9 y 2x~ ^'
^3y 4 a; 6 *
226
ELEMENTS OF ALGEBRA.
5_
12
24
12.
?^ = 16.
2/ 2a;
14,^
+  = _ 15.
l2y X
7.
y + l
__7^
~ 12'
12*
_ 5
~ 6'
= 2.
13.
14.
2 2/ 4a?
13^/^22;
71
'1
79.
15. <
5 16
 + —
X y
= 44.
y^x y
9.
ri5
8
17
2/
X ~
3
2
3
7
I y
a;
5
16.
11
2^
+
6
32/
17.
17
6 1/
—
5
a;
3
2*
10.
1
+
2 (2;  2) ' 3 (2 2/
3 5
1)
5 a; 10 4 (4 2/ 2)
= 5.
= 1.
11.
22/ 2
17.
2a;
2
5
4
"27
1
42/
1
11
"72
18.
SIMULTANEOUS SIMPLE EQUATIONS. 227
a; — 2 y + 2 ^g
3 1 1
^a; — 2 y +• 2 2
1 1 7
2x 3y""15*
a: — 5 v/ + 4 a: y
1
4 a; V
15
19. Suggestion. Reduce the first member of the second equa
tion to mixed expressions. Etc.
('2i _ 3y ^ 2 rL2. 25^86
l3^ 3y2a;+l 3'2lJ^2^
5, 2y _ * 25_1_6^
a;"^3y2a;+l la; y
94. In solving literal simultaneous equations, either of the pre
ceding methods of elimination may be applied, usually the method
by addition or subtraction is to be preferred.
Note. Numbers occupying like relations in the same problem, are generally
represented by the same letter distinguished by different subscript figures ; as,
«l ; «2 ; "8 > fitc* ? r«*d a one ; a ttoo ; a three ; etc.
They may also be represented by different euxents ; as, a'; a"; a'"; etc.;
read a prime; a second; a third; etc.
Kt AMPLE 1. Solve: 1^ ^
+ n
y
= a
(1)
hn,
iV
= «i
(2)
Process. Multiply (1) by
mj.
m^mx + m^ny = m^a
(3)
Multiply (2) by m.
m^mx + mn^y = ma
(4)
Subtract (4) from (3),
m^ny — mn^y = mjO 
w»«i,
or factoring,
(m^n — mn^)y — m^a 
 may
Dividing by mjn — muj,
^""h
 Oj m
^ nij^n 
mnj
Multiply (1) by rij.
mn^x f nn^y = n^a
(6)
Multiply (2) by n,
m^ux \ nn^y = na^
(6)
Subtract (6) from (5),
mn^x — m^ux = 71^0 
no,.
or factoring,
(mui — in^n)x = n^a 
 noj.
Therefore,
_a^n
m^n'
228
ELEMENTS OF ALGEBRA.
Example 2. Solve: ^
f X y
1^+5 + ^='"
x +
(1)
(2)
[ 2a6 ~ a2 + 62
Proceas. Free (2) from fractions, transpose, and factor,
{ahyx{a^hyy^Q (3)
Simplify (1), {ah)x^{a^lS)y = 2a(a+b) (ab) (4)
Multiply (4) by a ft, (abyx+(a^b'^)y = 2a(a+b) {aby (5)
Subtract (3) from (5), 2 a (a + 6) i/ = 2 a (a + 6) (a  &)2.
Divide by 2 a (a + 6), y = (ab)\
Substitute in (1),
Examples. Solve
a+b
+ ab = 2a. .'.x=(a + by.
f m n—'m(m+n)(b—y)__
J n{a\x) m(b~y) ~
! ryt
I — — + T = m2+n2
\^a + x b—y
Process. From (1),
Multiply (3) by ~,
+
n(a + x) m{b — y)
= m + n
+
a + X m^(b — y) m
— (m4n)
(2)
(3)
(4)
Subtract (4) from (2), ^3^  ^2Q,_y^
1
Simplifying,
Substitute
m{b — y) ^ m
m(by)
= 1 or
by
n
a + x
m2 in (2),
= n^ .. X
Example 4. Solve: <
xy +\
xy1
x + y + I
x + y
 a =
6 =
(1)
(2)
SIMULTANEOUS SIMPLE EQUATIONS. 229
Procesa. From (1), {a l)x  (a l)y = a + I (3)
From (2), (6 1):. + (6 l)y = 6 + 1 (4)
Divide (3) by a  1, x~y = ^—^^ (5)
Divide (4) by 6  1, x + y = ^3j (6)
2(a6l)
Add (5) and (6), 2 x = (q_i)(fc_i^ '
abl
•••^(al)(6l)
2(a6)
Subtract (5) from (6), 2 y = ^^. ^^^^.^^ •
ab
Exercise 86.
Solve :
 (ax\hy = m. ^ ( ax \ by = a^,
'\bx\ay = n. ' \hx \ ay = 1?.
^ nx + my = n. ^ (px'qy = r,
' \px\qy = r, ' \rx—py = q,
^ (ax = hy. ^(x + ay = ai.
' \bx\ay = c, ' \ax\aiy=l.
? + ?=!. (^ + ^ = a.
. J a b ab ^J^ \ m n
I bi aibi ^n ' m
5. <
3j/^2^^2 r_y X ^ 1
m ?i * ^ Ja{b a — b a + b
9y_6^^3 ' I ?/ , ^ ^ ^
230 ELEMENTS OF ALGEBRA.
= (a + h) y.
' \ cy + hx = a. ' \x + y = c.
rih; / J 14.  + 2' = 2.
l^ + r = cll+). {mx = ny
\^(m — n) y = {m {• n) X.
y X ^
a ai
(axhy
\(ah)x+ (a + b)y=2 {a^  l^).
22 (m{m7j) = n(x + ym).
\m (x — n — y) = n (x — n).
a b a fx + y+l^m + l
2g^a + a? by b 24. <^2/^+l ^"1
b a b \ X + y + 1 __ n+1
a + x b — y~a \y — x^l~l—n
25 / 3/  ^ + 2 (m  ?i) = 0.
■ 1 (a:; + 7i) (y + m) — (y — m) (x — n) = 2 (m — n)\
SIMULTANEOUS SIMPLE EQUATIONS.
231
95. Simultaneous equations with three or more unknown num
bers are solved by eliminating one of the unknown numbers from the
given equations ; then a second from the resulting equations ; and so
on, until finally there is but one equation with one unknown niunber.
Thus,
r 2 y + 2 + 2 y =  23 (1)
yf32 =  2 (2)
4a: + z=13 (3)
Example 1. Solve
3 + 3.
Process. Multiply (2) by 2,
Subtract (5) from (1),
Multiply (4) by 12,
Subtract (7) from (3),
Multiply (8) by 5,
Add (9) and (6),
Substitute in (4),
Substitute in (3),
Substitute in (2),
 20 (4)
2y + 6z =  4 (5)
52 + 2y =  19 (6)
4a; + 361; = 240 (7)
z36v= 253 (8)
52 180i;= 1265 (9)
178y= 1246. .. r =  7.
I 21 =20. .. x = 2.
12 + 2=13. .. 2=1.
y + 3 = 2. .. y = 5.
Proof. Substituting — 7 for y, 3 for x
f  23 = 
 2 = 
13=13
 20 =  :
(I), (2), (3), and (4), we have {
— 5 for y, and 1 for z in
23 (1),
^ ^^]' identities.
(3).
(4),
Kote. When the values of several unknown numbers are to be found, it is
necessary to have as many simultaneous equations as there are unknown
numbers.
EiLAMPLE 2. Solve:
J_ J 1__ 1
2z"^ 4y 32~ 4
1
1 4
0)
(2)
(3)
\2 ELEMENTS OF ALGEBRA.
Process. Multiply (1) by 2,  f
1 2 1
2y 3z2
(4)
Subtract (2) from (4),
5 2 1
6y~ 3z~2
(5)
Subtract (2) from (3),
2 4
I5y'^ z~^^
(6)
Multiply (5) by 6,
5 4
(7)
Add (6) and (7),
77 77
16 y~ 15
.'. y =
:1.
Substitute in (2),
1 1
.'. X
:3.
Substitute in (5),
fl 1
5 2 1
6 3z~ 2'
.'. z 
:2.
(1)
Example 3. Solve : 
1 1
ly + i"
Process.
s
(2)
(3)
Add (1), (2), and (3), ?+?+^ = a + & + c (4)
Divide (4) by 2, ^ + J + 1 = ^±A±f (5)
Subtract (3) from (5), ^ = ^±—
Subtract (2) from (5), ^ « + c  6
Subtract (1) from (5),
y 2
1 &4C — a
2
«+&c
2
y —
a6 + c
2
h\rC — a
SIMULTANEOUS SIMPLE EQUATIONS.
233
Exercise 87
Solve:
f'^x y+ z= 9.
I. ^ a:_2y+32= 14.
r4:x3y+2z = 40.
^<5xi9y7z = 4:7.
Ua;+8y32 = 97.
r2x32j+oz= 15.
3. < 32:+27/ z= 8.
V— a;+ 5y + 2s = 21.
rSxSy+ z= 0.
4.<2a;7y + 42:= 0.
v9 2:+5 7/+32= 28.
rx \ y \ z= 5.
6. ^ 3 7/5x + 72 = 75.
19 y 11 2+ 10= 0.
r.65// .95a: = .5.
6. < 5.1 a: 3.3 3 = 6.
V20.3 2 23.1a: = 21.
rax •\ hy '\ cz=iZ.
7. < rt x — 6 y + c 2 = 1.
Vaa: + 6y — • C2 = 1.
r.2a; + .ly + .32=14.
8 < .52:+.4y+.a2 = 32.
^.7y.8a:+.9c= 18.
9. <
2^ + 2 + 3 =
+2+1=
a; ?/
X y z
^ +  + '
6.
1,
17.
= 1.
= 1.
= 1.
11. <
234
ELEMENTS OF ALGEBRA.
fvhx + y + z = 14:.
\2v + x = 2y + z2.
14^^ ^ Sv  X + 2 y + 2 z = 19.
\ V X y z
^a — X h — y c — z
+ + = 0.
X
y
15. <
a — X h c _
X y z~
X y z
0.
15. Suggestion. Reduce fractions to mixed expressions. Etc.
'x\2y = 9,
16. ^3 7/ + 4^ =14.
72 + V = 5.
^2v + 52:== 8.
rx + y= 1.
\x \ z = b.
2 1_ 3
X y z
18. < ^   = 2.
z y
1 1_4
x^ z~ ^'
f4:y\Sx + z 2x + 2zy\l _ yz5
10
15
19. <
9?/ + 52:2;s 2y + a; — 3^_ 7x + z+S 1
r2 ^ 4 ~ 11 ^ 6 *
5^ + 32 2y+Sxz ^ ^ Sy + 2x+7
—4 12 + 2. = 0.1+ g
Queries. Upon what principle is elimination by addition and
subtraction performed ? What substitution ? What comparison ?
SIMULTANEOUS SIMPLE EQUATIONS.
235
Miscellaneous Exercise 88.
Solve :
'x+1 xl 6 r4a;+y = ll.
23
y
a(x^y) + h{xy) = l.
& (x + y) = 1.
7.
4.
5.
10.
X — a
= 0.
a b
0.
'a; + 2y=2324i;.
Sy + 2x = S — 4:z5v.
9v823 = 6a;7y.
^v = 25 4^ 16y 64a:.
■(m2 — n2) (5 ^^ 4. 3 y) = (4 ^ _ n) 2 m n.
„ a m vF
m^y
3 a; 15
/M
lf!=»
Ihh'
y ^
123
.!+l^
1=^
,ai bi '
{{m\n\'a)nx = n^y\{m\2n)mn.
11.
3 V + a: + 2 y  2 = 22.
4a; y432 = 35.
4v + 3a;2y = 19.
.21^ + 4^+2^ = 46.
(15 a; = 24 2 10 y + 41.
12. \ 15y=12a; I62+ 10.
ll8y(7 213)=14a;.
'Ul=z.
13. <
X
1
a; ■ z
y
2
+  = 11.
^U3.
y 2
236
ELEMENTS OF ALGEBRA.
14.
'x + y + z + v = 14:.
2x+Sy + 4:Z+5v = 54:.
4:x — 5y — 7z+9v = 10.
:Sx + 4:y + 2z3v = ll.
16.
) X + Z + V =
= 5.
10.
X + y+ V = 6.
x + y + z=12.
15.
18.
ax + by = 2m.
ax \ cz = 2 71.
,h y + c z = 6p,
rmx + ny + pz = m.
17. < mx — ny—pz = n,
\mx +py + nz = p.
Sx
2y , \ly
— + 1+ ^
10
45
4.X2
8 7
55^+ 71 y + 1
18
4 a; — 3 V + 5 45 ■
 + — 7
^lx =
v^2Z
17 + 2zStc.
20.
2(0 + 22/).
4.
y= 2.25\.75u—5v.
z = ll^u.
19. {u=ly^ I X
( ax +
\ay +
b X
by
cy = m.
cx = n.
X y m
21. < +  = .
\x z jp
I 1 1_ 1
\^z y n
22.
f :^+2/+2; = a+J+c.
i a+x = b\y = c+z.
9S 117^ 2(5lly) ^ 17.5 + 5y 312.5360a;
?jx'^ ll{yl)~ 32/ 36(a; + 5) ~
^3 4 1 ^^ fx ^ ^
^— + = 7.6. +l=4a;.
X by z
4
_ ., +  = 16.1.
^o X 2y z
+ l = 2y.
SIMULTANEOUS SIMPLE EQUATIONS.
237
1
y +
1
y~
26. <
a; —
X —
ii
y
lx = 0.
x\y = 2m^
27.
711 {li
mn
m+n
m—n\
mn
m—n
28.
m n V ^
 +  +  = 3.
X y z
m n P _ ^
X y"^ z~ '
2 m
X
n
0.
29.
30. ^
xy
x{ y
xz
= 70.
= 84.
X ■{ z
^ = 140.
y + z
31.^
xy
x+ y
yz
y + z
xz
^x \ z
xy
— ^^ = m.
32. ^
'ax\ by + cz = 0.
a^x + ly^y + (^z = 0.
.a^x + h^y'ic?^z = 0.
'a^z=:2.
x + y
xz
x+ z
yz _
^y + «
n.
{ax + a^y — a^yVc
2(3a;2y) _^_ bzy
34
35.
a; — 2g
3a; — 2i
m — n +
Sz7
n
2x — Sz
m
ly^ =
w
m + 71
2 7l6
7/1^ + 7?l 71 + 71^
238 ELEMENTS OF ALGEBRA.
CHAPTER XVII.
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS.
96. The solutions of the following problems lead to simultaneous
simple equations of two or more unknown numbers. In the solution
of such problems the conditions must be sufficient to give just as many
equations as there are unknown numbers to be determined.
Exercise 89.
1. If 5 be added to both numerator and denominator
of a fraction, its value is f ; and if 3 be subtracted from
both numerator and denominator, its value is ^. Find the
fraction.
Suggestion. Let x = the numerator,
and y — the denominator.
ra; + 5_ 3
— T~K — T'
By the conditions, •{ y ^
I a:3 _ 1
Solving these equations, x — 1, y =\\.
Therefore, the fraction is ^y
2. A certain fraction becomes equal to 3 when 9 is
added to its numerator, and equal to 2 when 2 is sub
tracted from its denominator. Find the fraction.
3. Find two fractions with numerators 5 and 3, respec
tively, whose sum is , and if their denominators are
interchanged their sum is f.
PROBLEMS. 239
4. A certain fraction becomes equal to § when the
denominator is increased by 3, and equal to J when the
numerator is diminished by 3. Find the fraction.
5. A fraction which is equal to  is increased to { when
a certain number is added to both its numerator and de
nominator, and is \ when 3 more than the same number
is subtracted from each. Find the fraction.
6. If a be added to the numerator of a certain fraction,
its value is a ; and if a be added to its denominator, its
value is ^ (a — 1). Find the fraction.
7. Find two numbers, such that two times the greater
added to one fifth the less is 36 ; three times the greater
subtracted from eight times the less, and the remainder
divided by 9, the quotient is 7.
8. Find two numbers, such that if the first be increased
by a, it will be m times the second, and if the second be
increased by 6, it will be n times the first.
9. Find two numbers, such that if to J of the sum you
add 18, the result will be 21 ; and if from  their differ
ence you subtract , the remainder is 3.65.
10. A farmer sold to one person 25 bushels of corn and
52 bushels of oats for S 38.30 ; to another person 42 bush
els of com, and 37 bushels of oats for $35.80. Find the
number of dollars per bushel received for each.
11. A farmer sold a bushels of corn and b bushels of
oats for m dollars ; also at the same time, c bushels of com
and d bushels of oats for n dollars. Find the number of
dollars per bushel received. Apply the result to 10.
240 ELEMENTS OF ALGEBRA.
12. A grocer bought a certain number of eggs, part at
2 for 3 cents and the rest at 5 for 8 cents, paying $7.50
for the whole. He sold them at 23f cents a dozen, and
made $2 by the transaction. How many of each kind did
he buy ?
13. A grocer bought a certain number of eggs, part at
the rate of a eggs for m cents and the rest at the rate of h
eggs for n cents, and paid c dollars for the whole. He
sold them at d cents a dozen, and made ^ dollars by the
transaction. How many of each kind did he buy ? Apply
the result to 12.
14. A number is expressed by three digits. The sum of
the digits is 8 ; the sum of the first and second exceeds
the third by 4; and if 99 be added to the number, the
digit in the units' and hundreds' place will be inter
changed, rind the numbers.
Suggestion. Let 2 = the digit in units' place,
and y = the digit in tens' place,
also X — the digit in hundreds' place.
Hence, 100a:+ 10y + 2 = the number,
and 100 2+ 10^ + a: = the number with the digit in units'
and hundreds' place interchanged.
By the conditions,
X + 2^ + z = 8,
a: + 1/ — 4 = 2,
100 a: + 10 y + 2 + 99 = 100 z + 10 2/ + a:.
Solving these equations, z — % y = 5, x — 1.
Therefore, the number is 152.
Note 1. In verifying, the results should be tested directly by the conditions
of the problem. Thus, in the above, the sum of 2, 5, and 1 is, as one condition
requires, 8. The sum of 1 and 5 exceeds 2 by 4» The sum of 162 and 99 is 251
fiB required.
PROBLEMS. 241
15. A number is expressed by three digits. The middle
digit is twice the left hand digit, and one less than the
right hand digit. If 297 be added to the number, the
order of tlie digits will be reversed. Find the number.
16. A number is expressed by three digits. The sum
of the digits is 18 ; the number is equal to 99 times the
sum of the first and third digits, and if 693 be subtracted
from the number, the digit in the units' and hundreds'
place will be interchanged. Find the number.
17. The sum of the three digits of a number is n ; the
number is equal to a times the sum of the first and third
digits, and if m be subtracted from the number, the digit
in the \mits' and hundreds' place will be interchanged.
Find the number.
18. If a certain number be divided by the sum of its
two digits the quotient is 3, and the remainder 3 ; if the
digits be interchanged, and the resulting number be di
vided by the sum of the digits, the quotient is 7, and the
remainder 9. Find the number.
19. If a certain number be divided by the sum of its
two digits the quotient is a, and the remainder b ; if the
digits be interchanged, and the resulting number be di
vided by the sum of the digits, the quotient is c, and the
remainder m. Find the number.
20. The sum of the three digits of a number is 16. If
the number be divided by the sum of its hundreds' and
units' digits the quotient is 77 and the remainder 6 ; and
if it be divided hy the number expressed by its two right
hand digits, the quotient is 16 and the remainder 5. Find
the number.
242 ELEMENTS OF ALGEBRA.
21. The sum of the three digits of a number is 9. If
the number be divided by the difference of its hundreds'
and units' digits, the quotient is 157, and the remainder 1;
and if it be divided by the number expressed by its two
righthand digits, the quotient is 21. Find the number.
22. A, B, and C can together do a piece of work in 12
days ; A and B can together do it in 20 days ; B and C
can together do it in 15 days. Find the time in which
each can do the work.
Suggestion. Let x = the number of days in. which A can do it,
and y = the number of days in which B can do it,
also 2 = the number of days in which C can do it.
1111111 111
Theequationsare + ^ + ~ = 12' ^ + 2^= 20' ^^^ y + z = \b'^
from which a; = 60 and y = 2 = 30.
23. A and B can do a piece of work together in 48
days ; A and C in 30 days ; B and C in 26 days. How
many days will it take each, and how many altogether, to
doit?
24. A and B can do a piece of work together in a days;
but if A had worked m times as fast, and B n times as
fast, they would have finished it in c days. How many
days will it take each to do it ?
25. A drawer will hold 24 arithmetics and 20 algebras;
6 arithmetics and 14 algebras will fill half of it. How
many of each will it hold ?
26. A purse holds 19 crowns and 6 guineas ; 4 crowns
and 5 guineas fill JJ of it. How many will it hold of
each ?
PROBLEMS. 243
27. A purse holds c crowns and a guineas ; ci crowns
tn
and ai guineas will fill — th of it. How many will it bold
of each ?
28. A and B together could have completed a piece of
work in 15 days, but after laboring together 6 days, A was
left to finish it alone, which he did in 30 days. In how
many days could each have performed the work alone ?
29. Two persons, A and B, could finish a piece of work
in m days; they worked together a days when B was
called off and A finished it in n days. In how many days
could each do it ?
30. A can row 8 miles in 40 minutes down stream, and
14 miles in 1 hour and 45 minutes against the stream.
Find the number of miles per hour that the stream flows,
also that A rows in still water.
Suggestion. Let x = the number of miles per hour that A can
row in still water,
and y = the number of miles per hour that the
stream flows.
Then, x + y = the number of miles per hour that A can
row down the stream,
and X — y = the number of miles per hour that he can
row up the stream.
Since the distance divided by the rate will give the time, by the
conditions,
8 2
x + y~ 3*
31. A can row m miles in h hours down stream, and mi
miles in ^i hours against the stream. Find the number of
miles per hour that the stream flows, also that A rows in
still water. Apply the result to problem 30.
244 ELEMENTS OF ALGEBRA.
32. A boatman sculls down a stream, which runs at the
rate of 5 miles an hour, for a certain distance in 3 hours,
and finds that it takes him 13 hours to return. Find the
distance sculled down stream, and his rate of rowing in
still water.
33. A man who can row at the rate of 15 miles an hour
down stream, finds that it takes 3 times as long to come
up the stream as to go down. Find the number of miles
per hour that the stream flows.
34. A waterman rows 30 miles and back in 12 hours ;
and he finds that he can row 3 miles against the stream
in the same time as 5 miles with it. Fiud the number of
hours in going and coming respectively ; also, the number
of miles per hour of the stream.
35. A waterman can row down stream a distance of m
miles and back again in h hours ; and he finds that he can
row h miles against the stream in the same time he rows
a miles with it. Find the number of hours in going and
coming, respectively ; also the number of miles per hour
of the stream, and his rate of rowing in still water.
36. Five pounds of sugar and 3 pounds of tea cost
$2.05, but if the price of sugar was to rise 40 %, and the
price of tea 20 % they would cost $2.51. Find the num
ber of cents in the cost of a pound of each.
37. If / pounds of sugar and h pounds of tea cost m
dollars, and the price of sugar was to rise a % , and the
price of tea h %, they would cost n dollars. Find the num
ber of cents in the cost of a pound of each.
PROBLEMS. 245
38. The amount of a sum of money, at simple interest,
for 11 months is S1055; and for 17 months it is S1085.
Find the sum and the rate per cent of interest.
39. The amount of a sum of money, at simple interest,
for VI months is a dollars ; and for n months it is h dollars.
Find the sum and the rate of interest.
40. A grocer mixes three kinds of coffee. He can sell
a mixture containing 2 pounds of the first kind, 9 pounds
of the second, and 5 pounds of the third, at 18 cents per
pound ; or one composed of 6 pounds of the first, 6 pounds
of the second, and 9 pounds of the third, at 19 cents per
pound ; or one composed of 5 pounds of tlie firet kind, 2
pounds of the second, and 18 pounds of the third, at 22
cents per pound. Find the number of cents in the cost of
a pound of each kind.
41. The forewheel of a carriage makes 6 revolutions
more than the hindwheel in going 120 yards ; if the cir
cumference of the foie wheel be increased by J of its pres
ent size, and the circumference of the hindwheel by J of
its present size, the will be changed to 4. Find the
number of yards in the circumference of each wheel.
42. The forewheel of a carriage makes a revolutions
more than the hiudwheel in going b feet. If the circum
ference of the forewheel be incieased by — th of itself, and
8 ^
that of the hindwheel by  th of itself, the hindwheel
r
will make c revolutions more than the forewheel. Find
the circumference of each wheeL
246 ELEMENTS OF ALGEBRA.
43. A grocer has two kinds of coffee. He sells a pounds
of the first kind, and h pounds of the second, for m dollars;
or, ax pounds of the first kind, and hi pounds of the second,
for mi dollars. Find the number of dollars in the price of
a pound of each kind.
44. A jeweller has two silver cups, and for the two a
single cover worth 90 cents. If he puts the cover upon
the first cup it will be worth 1^ times as much as the
other ; if he puts it upon the second cup it will be worth
lyig times as much as the first. How many dollars in the
value of each cup ?
45. A jeweller has two silver cups, and for the two a
single cover worth a dollars. If he puts the cover upon
the first cup, it will be worth m times as much as the
other ; if he puts it upon the second cup it will be worth
n times as much as the first. How many dollars in the
value of each cup ?
46. A broker invests $5000 in 3's, $4000 in 4's, and
has an income from both investments of $315.50. If his
investment had been $1000 more in the 3's, and less in
the 4's, his income would have been $5.50 greater. Find
the market value of each class of bonds.
Note 2. 3's means bonds which bear 3 % interest. The " quoted " price of
a bond is its market value. Thus, a bond quoted at 115i means that a $100
bond can be bought for $115.50 in the market.
47. A broker invests m dollars in a's, n dollars in c's,
and has an income from both investments of h dollars. If
his investment had been d dollars less in the a's, and more
in the c's, his income would have been p dollars less. Find
the price paid for each kind of bonds.
PROBLEMS. 247
«
48. A and B do a piece of work together in 30 days,
for which they are to receive $1G0. But A is idle 8 days
and B is idle 4 days, in consequence of which the work
occupies 5J days more than it would otherwise have done.
Find the number of dollars received by each.
49. A and B do a piece of work together in m days, for
which they are to receive c dollars. But A is idle a days
and B is idle h days, in consequence of which the work
occupies n days more than it would otherwise have done.
Find the number of dollars received by each.
50. The amount of a sum of money, at simple interest,
for 5 years is S600; and for 8 years it is $660. Find the
number of dollars in the sum, and the rate of interest.
51. The amount of a sum of money, at simple interest,
for a years is m dollars ; and for h years it is n dollars.
Find the number of dollars in the sum, and the rate of
interest.
52. If a grocer sells a box of tea at 30 cts. a pound, he
will make SI, but if he sells it at 22 cts. a pound, he will
lose S3. Find the number of pounds in the box, and the
number of cents in the cost of a pound.
53. The smaller of two numbers divided by the larger
is .21, with a remainder .04162. The greater divided by
the smaller is 4, with .742 for a remainder. Find the
numbers.
54. The smaller of two numbers divided by the larger
is a, with a remainder m. The greater divided by the
smaller is h, with c ibr a remainder. Find the numbers.
248 ELEMENTS OF ALGEBRA.
CHAPTEK XVIII.
EXPONENTS.
97. An Exponent is a figure or term written at the right
of and above a number or term (Art. 21).
'^ m
Thus, in the expressions 5^, a% 6", and (a + by; 2, c, —, and 3
are exponents.
Zero Exponents. When the dividend and divisor are
equal the quotient is 1.
Thus, ^,= 1; ,= 1; ~,= i; ~ = I; etc.
But (Art. 30), 32 = 322 = 30; ^ = aO; ^^ = «^ ^ = «^ etc.
Therefore, it follows that a^ = I. Hence, in general,
I. Any expression with zero for an exponent is 1.
The Reciprocal of a number is unity divided by that
number.
Thus, the reciprocal of n is ; of n + m is
n \ m
Negative Integral Exponents.
a3 X a8=a83 = «o^i
Divide by a^,
a» X a« = «"« = a^= \
Divide by a",
a~** = — . Hence, in
EXPONENTS. 249
II. A negative integral exponent indicates the reciprocal
of the expression with a corrcspoiuling positive exponent.
The expression a", where n is any positive integer, represents tht
product o/n equal factors, each equal to a. It has been shown that :
Art. 21, a^ X a* — a*»+".
Art. 30, a"^ — a^ = a"*", where m is greater than n.
Art. 30, ti^ T a'*= ^_,^ , where m is less than n.
Art. 27, (a*")" = a*"", whatever the value of m.
Thus,
By Art. 21, a* X a" X a' x .. . . a"* = a»+«+ ''+••'».
Take n factors of a*, a*, a**, a"*, and suppose each of the n ex
ponents equal to m, then it follows that
(amy _ Qtnn^ Hencc, m can
be positive or negative, i itegral or fractional.
By II.. <'" = ii
a"*
Multiply by a"*, a" X a"» = — •
If m is greater than n, Art. 30, ;; = a*"".
Therefore, a" >^ a*" = a"**.
If m is less than n,
a* 1
a» a
i« — w
By II., 7=^ = «'
a"
Therefore, a» x a*" = «"•'• for all possible integral
values of m and n.
98. By Art. 27, (ai)" = a" X 6".
Therefore, o^*" = (aft)*.
Similarly, a" X ft" X c" X . . . . p" = (a 6 c . . . . />)■
250 ELEMENTS OF ALGEBRA.
If n is a negative integer,
«" X *" = a« X 6" =(«6)» =("*)•
Similarly,
a« X i^ X c» X ....jD"= {ahc ...
general.
. Py
Hence, in
I. The product of tv:o or more factors, each affected with
the same exponent, is the same as their product affected wiih
the exponent.
By IL, Art. 97, a" f &« = a« 6».
Also, a« 6« = (a 6 1)« = (? ) •
Therefore, a** f ft" = [ r
Similarly, a" ^ &»» = f r) Hence, in general,
II. TAe quotient of any two factors, each affected with
the same exponent, is the same as their quotient affected with
the exponent.
Illustrations: I. 22 X 3^ = (2 X 3)2 = (6)2 = 36 ; 28 X 3^ X 48
= (2 X 3 X 4)8 = (24)8 = 13824 ; 22« X 32« X 42" = (2 X 3 X 4)2 «
= [(24)2]" ^ (576)«= ^ ; (f)2 X (f)=^X (i)2z. ( x f X i)2
1
(}) = ^=16.
16\8
II. 242^62= (5^4)2= (4)2:^16; (_16)8(4)8=(^)
1\4ot /'1\4'» 1
These examples are said to be simplified, that is, they are expressed
in their simplest forms.
EXPONENTS. 261
Exercise 90.
Simplify :
1. (n^f X {a^f X (71)2; (J)2 X (2)2 X (§)2  (if.
2. (a;* y'")8 f (oj V")^ (216 2^2)4^(54 2^2)4^
3. (a)8 X (aj2)8; (^ri)" X (^■); (a;)* x (2:"^)*.
4. (x)" X {a^Y; (f)" X (})" X (2)""; {aHf X (a68)6
5. (2 71)10 X (2 im)iO; (a 6ic2)3  (aifeac*?/^)^.
6. (4a*^a:«)"^(22a3*2;V)"'*; (^iy*)^^(^^rV.
7. a X (3 2>"')" X (ci)™; (:r)i  (^'y".
8. («2 5)2 X (« ?>3)2; (rtS J8 + ^6) 3 ^ (a6an3)8
9. (i)" X {^T X (J)" ; (a2'' + a" ?»2'')i x («"  &2«)i.
10. («i)^ X (xi)5 X (xt)6 X (at) 5 X (&i)^
11. (^^"+*)'' X (i2*j" X (an)" X {bnY; aS ^ (2 a)8
12. « 2x(2a)2f.^^y; (2«2)2x^^y'x(a)2.
13. (ai V^x)' X (.r2v^6)8. (,,i)2'' X (2^)2" X (c")2".
14. (2")" X (2"!)'' X (22—1)— X (22"+i>'» X (r)\
15. (2''+i)"' X (2— "+")"• X (2"'!)'" X (4"^)'" T(ir))'".
16. [(2:y)8]X[(rr + 7/)]8; ()" x ff)  (J)
252 ELEMENTS OP ALGEBRA.
99. Positive Fractional Exponents. If m and n are both
positive integers,
Kan) = a"*.
m
Take the nth root of both members, a'* = \/a^.
m
Therefore, a« means the nth root of the mth power of a, or the
mth. power of the nth root of a. Hence,
The numerator, in a fractional exponent, denotes a power,
and the denominator a root.
The denominator of the exponent corresponds to the index of the
root. Thus, (81)1 = \/{Siy = (^8iy = (3)3 = 27.
m
In a« = /y/a"*, m is the index of the power, and n is the index of
the root ; also a, m, and n may be any numbers. The expression
may be raised to the power indicated by the numerator of the expo
nent and then extract the root of the result indicated by the denomi
nator; or, extract the root first and then raise the result to the power
indicated by the numerator of the exponent. Thus,
(8)1  V^FS? = v'64 = 4 ; or, ( 8)f = (v^^)' = {~ ^Y = 4
Notes: 1. a« is read "a exponent —n;" a" is read "a exponent ;
a~ « is read "a exponent ." These are abbreviated forms for "a with an
exponent —n; etc.
m
2. It is manifestly incorrect to read a« " the  th power of a." There is
no such thing as a fractional power.
3. We must be careful to notice the difference between the signification of a
fraction used as an exponent, and its common signification. Thus, f used as
an exponent signifies that a number is resolved into five equal factors, and tlie
product of four of them taken.
mXc mc
100. By Art. 73, a»» = a« x " = anc;
m
e
m 7HTC n
also, a** = a" ^ " = a^ Hence,
EXPONENTS. 263
L Multiplying or dividing the terms of a fractional ex
ponent hy the same number will not change the value of the
expression.
^n^a^"^.
But
a^^^=^a^,
and
^a^=^/^a.
Therefore,
Jrn  \/;^a.
Hence, in general,
II. The mnth root of a number is equal to the mth root
of the nth root of that number.
niuBtrations.
2* = 2* ; 6» = 6i ; 62« = 6«^; ^^64 = '^ ^64 = ^ = 2.
101. Negative Fractional Exponents. If m and ti are
both positive integers,
( Y
\a »•/ = a""*.
By II., Art 97, a« = ^.
Take the nth root of both members,
m 1
a » = — . Hence,
Ajiy expression affected with a negative fractional expo
nent is equal to the rccipror/U of the expression with a cor
responding positive exponent.
_"* 1 "• 1
Notes : 1 . From the relation a »• = ^ , a" = — ^ . Hence, the method of
o* a *
Art. 30 is true for fractional exponents.
2. Any factor of the dividend may be removed to the divisor (or from the
numerator to the denominator of a fraction), or any factor of the divisor to
the dividend, hy changing the sign of its exi>onent.
254 ELEMENTS OF ALGEBRA.
Illustrations. 2^^ = ^ = ^^1 (1)"^ = (^ =  = I' ^ = «"^
,,.1 3 1 1 1 2* X 3 3 / u^
X (i) ' X 4 ' = ^, X I X 3 = j,~ =  ti ^x")~
1 11:^,
^ ^  x^ ' x~ x^^ •
102. (ah^y=ab.
Take the nth root of both members,
J 1 1 1 j^
Similarly,* an x &" X c« X . . . . ;?" = (a 6 c . . . . ;?>. Hence,
I%e product of two or more factor's each affected with the
sa7ne root index, is the same as their product affected with the
root index.
In the same manner we can prove that
Kotes : * 1. If we suppose that there are m factors oi a,h, c, p, and that
each factor is equal to a, then it follows that
By Art. 99, [a^Jn = an.
Therefore, Va»/ = an.
2. Similarly, \an) = a. Hence,
Tfie nth power of the nth root of a number is equal to that number.
Illustrations. (A)* X (f,)^ X 8* = (f X  8)* = ^'^ = ;
EXPONENTS. 255
»v
103. (a" X a'T' = (j««+»*.
m b JL
Take the ncth root, a« X a« = (0™'+"*)
m . 6
By Art. 99, (a"» «+"«>)"'' = a" ''^ «.
m 6 m 6
Therefore, a»Xae = a«'^«. •
iw ft r t ?j.*j.'"i. ?
Similarly, a* X a« X a» X • • a« = a» « i"*""" «. Hence,
I. Th£ product of several expressions consisting of the
same factory affected with any exponent, is the factor with
an exponent equal to the sum of the exponents of the factors.
By Arts. 101, 21, c* f ac = a* X a « = a»» «. Hence,
II. The quotient of two expressions coTisistiiig of the same
factory affected with any exponent, is the factor with an
exponent equal to that of tlic divideiid mimis tJiat of the
lUuBtrationa : I. 5* X 5"^ X 5 = 5*"*"^* = 5* = >^125 ;
X* X a:* X a:" = x"+*.
II. 2*r2* = 2^"* = 2i = v^2; (a + 6)' ^ (a+ 6)* = (« + 6)'"*
= (0 + 6) A.
Exercise 91.
Simplify :
1. 16f X 16i; 25i X 25i; 3^ x (^; a"! X ^.
n n 6_»» X 2 m
2. aixa^Xa^; n^' X n '; m "Xw ''; 2iV2.
3. y"^ X 7/" » ; a' T a~'; rt* X J; (a^)* r (a^)h
4. (2)i(32H; a* at; (^J  (..y)i.
256 ELEMENTS OF ALGEBRA.
5. x^ ^ rr2«; a '^^m ^ ; (a h)^  (ah + hi)i.
6. 32« ^ 3" ; (a  &)" ^(ab); (x^/2)h ^ (// 2;^)?.
\a^b ^J \a^b^J \ax^ J \x ^ J
8. {a 2 xf x{a2xf x{2xafx{ci2 x^.
10. (.r + yy'' ^{x + ^)"; a3^+2y _^ a2^3^; lA^rA.
/ m \ ^ / »« — 1 \ —1— / \ "' + ^" / \ ^
11 1*^ Xmnp / 3? \mn/) f CC\ n f CC\m
13. at = ai; 2" x (2»)»' x 2" + ' X 2»i x 4".
^^2 . i^y {off (x'f . , ; .. , ^ /^V''"
X04. caa"'=(a")=a.
Take the n qih. root of the first and last members,
(m r ( \p mp rp tp
The principles of this chapter are true, whatever the values of
a,h,c,....m, n, p, and q ; that is, a,b,c, m, n, p, and q can be
positive or negative, integral or fractional.
EXPONENTS. 257
niustrations. (2' X 3* X 4"^)' = s'""* X S*""' X 4*'** = 2«
X 2* X 3* X 41 = ^2 X V3 ; • L(a"^)"«]fl^ r I [(aW^JJlJ
N m
a"« »• = a* — a = a.
105. Negative and Fractional Root Indices.
_ j>i_ _w \ 1
an V«
m _m
_ e _£ £ 1 1
Similarly, y^a™ = a»« = a « = — = "^TZ* Hence,
^ negative root index, either integral or fractional^ indi
cates iJie reciprocal of Uie expression with a corresponding
positive index.
Note. Since it is impossible to extract a fractional or negative root, or raise
an expression to a fractional or negative power, in order to perform the opera
tion indicated by such indices some preliminary transformations must be made.
lUuBtrationB. ~i/^ = : — =  = r = —^ ; i/4a« = (4 ay
_ I _ ± ^_ _ ± _ J = 1
a*
Exercise 92.
Simplify :
1. 1^27; V^; /32 miO; Vsla^; V^.
2. 1^8; [(63)2(a*)8(68)(a6Ji)2]6; ^8a*6— ic"2
17
258 ELEMENTS OF ALGEBRA.
2
V ^ / \y'J Kni^n^J ' vo«'"^'*^"^' V25'
1 1^ / J_ \a2  62
Queries. What does a negative exponent indicate 1 A fractional
exponent ? A negative fractional root index ? Any expression with
for its exponent = ? Why ? What is the product of as and a^ 1
Prove it.
Miscellaneous Exercise 93.
Express with fractional exponents and negative power
indices :
1. ~\^^; ~^; 4"^; 'V^; (^a)^; ^^5^2
m
Express with radical signs and negative integral root
indices :
m O
3. at; a^hic~^; 4:ah~^; 7a~^x~''; —  .
X 4
EXPONENTS. 259
Express with radical signs and fractional root indices :
n n
9 1 m m X T7"«
4. at; (4a2)!; alz^U; a'"5"; a" ft";
xy.
Express with fmctional exponents and fractional power
indices :
5. ^Jbi; y/2'6; 3v'(8a8); ^a'~^; Va; ^^5.
Express in the form of integral expressions :
Sa^b 5 m* a x~^ x~^ a^
c2 ' ahc' nr 4^1' 4^^' ^' aibl
if
Express with literal factors transposed from the numera
tors to the denominators:
Simplify and express with positive exponents :
8. 4^; v^^^a. yj(h^; "^/^; «»; [Va^ H V^]"".
_ 2 ai X 3 ai a 2^ x a'^ x v"^ «/ ^ s.y
9. == ; —=r. — ^^ ; Vm3 ^ V7/il
10. 2;ix2a;i; (^V*; a^ x «i X aJ.
11. ^4^; (j^y^; V^^h X ^J'^Fs
12. aUiaxai6Ui; (f^)"'; y/^
260 ELEMENTS OP ALGEBRA.
13. aH^c^ XaH^ci; \l^\ i/(^~H*)^.
14. y" X y X "; (x + y)^ X n^ X n~r^ X Vti.
15. Y/(m^'^V'; y/«»6"i i()"'; tF^.
16. (77ii\^a)3 X 'v/(a2 V^"; y^^' + —
17. VaF^W^'^{ah~~^y; l^^ll^.
2"' 2»2
a; '■m "^ n
^^ m2 ^ mw Ml , J' (m27^2)2
2"(2"i)" 1 2" + ^ 4""^^
2»+i X 2"" 4^"' (22»)"i • (2"i)" + >
(9"x32x5kV27»
V / / ; (2" X S"*)"" Is" X 6*"")"'
23. pir^^
EXPONENTS. 261
Multiply :
24 a*fe~i  a^h"^* + 1 by a^h~^' + 1.
25. a^ + a2»&*' + h^' by a" — a^"^)*"* + ^'.
26. a^h '—a* 5 •+« *7>i— a »6« by a«6 i + a "ft*.
Divide :
27. a^ + rt^ M"' + &• by a» + a2^ M"' + ft*".,
28. a,*"*"*) — y2m(ml) ^y ^{nl)^^(m«
29. a;**"*  if^* by ar"*"'' + i/^'"^
30. a;^"'*"  ^m'am by ^"^^"^ ± f*'*,
31. a^ —3^+4:a*'*x*'—4:a^x^' by a2* + 2a^2:<'' — .r^*.
32. a3+a~i* by a5 + rt^; riT/i + mx^ by n^yi + w^a;^.
Separate into two factors :
33. a^b; a'^  ftf ; aV  62«.
Expand :
34. (aU6ia:)*; (a:2a;i)8; [(a"^ «?)']".
Resolve into prime factors, and find the products of:
35. N?'!^, 4^2, \/96, \^
36. ^12, v^72, \^, ^, ^^^576, V2l
262 ELEMENTS OF ALGEBRA.
Find the cube roots of :
38. 8 a2  12 a V + 6 af  a"!
39. a;3__9^+27a:i27ar3.
Find the 6th roots of :
40. a;« + ^  6 (a;* + ^) + 15 (^x^ + i)  20.
41. 729  2916 a2« + 4860 «*"  4320 a^" + 2160 a^"
576^10" + 64a^2«,
42. a; 12  6 :r 10 + 1 5 a; 8  20 ^^ 6 + 1 5 ic 4  6 aj 2 + 1.
Simplify and express with positive exponent :
!i+i, ifan i/4 X 4"i
H:i,
y/4«i >^ 4„ + i
4fi ^"^"^ [(8a6?>)2"]5" Q
9(a;0 + 2/0 + ^)2m3' [(4a3 6)5"]2''' ^ + T •
.,, (20a3H8a;2?/212y^)" (m^ + ^^)^ (^^  n^)^
[4 (2:2 + 2/2).f ' ' m6?i6
RADICAL EXPRESSIONS. 263
CHAPTEE XIX.
RADICAL EXPRESSIONS.
106. A Surd is an indicated root that cannot be exactly
obtained ; as, V5 ; 'V^f ; \^a^.
The Order of a surd is indicated by the root index.
Surds are said to be of the second^ third y fourth, etc., or nth order,
according as the second, third, fourth, etc., or nth roots are retjuired.
Thus, 'v/a, ^a, \/b, etc., yx, are quadratic, cubic, biquadratic,
etc.
Surds are of the same order when they have the same root index ;
as, ^b, ^a\ and ^¥.
A surd is in its simplest form when the expression un
der the radical sign is integral, and in the lowest degree
possible ; as, ^32 a* = \/2^ a^ x 4 a = 2 a v^4 a.
Similar or Like Surds are those which, when reduced to
their simplest forms, have the mme surd factor ; as, 3 \/3
and a/3 ; 2 a vh and c ^/h. Otherwise the surds are
dissimilar.
Hotes : 1 . When a surd is expressed by means of the radical sign, it is
called a Badical ExpressioxL
2. An Irrational Expression is one which involves a surd ; as, V3 ;
a \h \c^.
3. An indicated root may have the form of a surd, without really being a
8uixl. Tims, Vi and Va» have the f(rrm of surds.
4. Rational factors or expressions are those which are not surds ; as, 2;
a*x — bf^y.
5. Since a" — a^P, surds of the form Va^ and fo^ are equivalent sards
of different orders.
264 ELEMENTS OF ALGEBRA.
6. A Mixed Surd is the product of a rational factor and a surd factor ; as,
a V6 ; 3 Vb.
7. An Entire Surd is one in which there is no rational factor outside of
the radical sign; as, V2; \'a^; Vx.
8. A binomial surd has two terms, and involves one or two surds; as,
a \b Vx] a Vx — b yy A compound surd or polynomial has two or more
2 3  4 
terms, and involves one or more surds ; as, y2 + 3 4/4 — 5 V3 ;
a\h c + 2Va.
9. Quadratic surds are of most frequent occurrence.
107. The methods for operating with surds follow from an appli
cation of the principles of Chapter XVIII. Thus,
f = V^f . 2 a2 &3 == ^(2^2y3y3 ^ ^^^;^9; j^ general,
n
a = a^ = a^— ^a^. Hence,
I. To Reduce a Rational Factor to the Form of a Surd of
any Order. Raise it to the power indicated by the root index, and
place it under the radical sign.
2V'3 = V2' X 3 = Vl^. f ^9 r= ^(1)3 X 9 = ^. In general,
n \ 1
a ^x = an xn — (a'*;r)» = y'a^. Hence,
II. To Change a Mixed Surd to the Form of an Entire Surd.
Reduce the rational factor to the form of the surd, multiply by the
surd factor, and place the product under the radical sign.
V72 = V62 X 2 = 6 V2. ^1029 a* = (7^a^ X Says = 7 a ^3a.
9 3/7 _ 9 i3/iZi _ ?Jl1 ,3/T ^.VJI _ «,V 1 X^'«
^Vu^y2x4~ 2  V4. 2V2a3  2V2a3 X 23a
a / Sa ^j —
"^ 2 V 2^ = i V 8 a In general ,
Hence,
RADICAL EXPRESSIONS. 265
III. To Reduce a Snrd to its Simplest Form, ii tiie surd is
integral, remove from under the radical sij^n all factors of which the
indicated root can be exactly obtained.
If the surd is fractional, multiply its numerator and denominator
by such expression that the indicated root of the denominator can be
exactly obtaiued.
\/2^ a X v^o^ = \^'2^a x a^ = a \/2. In general,
_ — — A J * * *
^a X ^/b X ^c X . . . . ^ = a'* X b" X C^ X . . . . p^ = (abc . . . . py
= \^a be p. Hence,
IV. To Find the Product of Two or More Surds of the Same
Order. Take the product of the expressions under the radical signs*
and retain the root index.
In general,
^/^^\/r=(^y=\/'f Hence,
V. To Find the Quotient of Two Surds of the Same Order.
Take the quotient of the expressions under the radical signs and
retain thn root index.
f/i5^64 = ^64 = 2. (/ V25^ = 1^(2»)'' = 2^ = 4. In general,
^'^ = (""j* = a^ = "^a. Hence,
VI. To Find the vith. Root of the lith Root of an Expres
sion. Take the mnth root of the expression.
Note. It is sometimes easier to j>erfonn operations with .simls if the arith
metical numbers contained in the surds be expressed in their prime factorSf and
fractional exponents be used instead of radical signs,
266 ELEMENTS OF ALGEBRA.
Exercise 94.
Express in the form of surds of the 3d and nth orders,
respectively :
1. 1; ; 22; 4"; 2 a"; Sahc; S x; a^; of; af'y\
Express as entire surds :
2. JV2; 1^3; 5 V32 ; f V^; leVflf; abVbi.
3. a 4/d^ b 68 ; 3 a^ ^ofc^ ^ i '^^ ; 2x</J^;  ^.
5. 5;.^^25^i; (,»l)v/^; '^^±^\I^^EI .
' ^ m — 1 771 — 71^ m + n
^' ;r^V^r^' ^V^?^' ^"V^r; ^^Vs^'
Express in their simplest forms :
8. ^288; 3V150; •^^^^IIS?; fV90; 2 a^W^.
9. V3i; ^Jl; ^J^; ^'1029; Vf; V^ ; ^
10. <1\ ;J; ^=T08^a, ^3^i;;i5ro, ^Z'^
11. ^^?n=^^\ ^'V^; !^v/— •
RADICAL EXPRESSIONS. 267
. ^', ^7290a3j6m^2. ^^J^; ^a^^+Y"
12. ,/
^586
13. V(a; + y) (a^  ^) ; Va«2  8 aa; 4 16 a.
14. ^^yJtlEI^LlIl, ^ 1715 ^^V^
Simplify :
15. Vl2 X Vl8 X V24; V54  Vl) V^Vlf.
16. ^Fex ^^54x^/128; [v0[28^ h ^5^6^] ^ >^9^.
17. ^v^^:^. X
V50a8 66^ V32a63
18. V2^ a8 ^6 X vOUe a2 m2 a:^ X v^56 a^ m^ x\
19. (^53a«fe9 4 ^25 a* 6^) x ^125 aH X <^W^.
20. (V6M ^ V63~?) X v^54^ T v/feT: \/'^^^^^.
21. (^iiT?^ X ^aiftic) ^'(v^a:^oyo x ^^lO^X
22. (^^^)^V20736; 'i^ivF^  aX^.
23. Vf a8 X Vf a2 X V.f ai X V2.5 a"*.
24. \1\J </W^^^\ (16 aH2)i X (ai h^f r ^2 J h.
108. ^5/2 = •^^21^ =^8. 3^=3*^]^2^M? = 3]!J^l6.
In general,
p pxw
y^aP = «" (n > /)) = a* »< "• = "^aP"*. Hence,
268 ELEMENTS OF ALGEBRA.
I. To Reduce a Surd, in its Simplest Form, to an Equivalent
Surd of a Different Order. Divide the required root index by the
root index of the surd, and multiply the power and root index by the
quotient.
TheL.C.M. oftheroot
indices (3, 9, 6) is 18.
In general,
pm
Pin
^6w = fe"*" (m > pi) =: "^6p.». Hence,
II. To Reduce Surds, in their Simplest Forms, to Equiva
lent Surds of the Same Lowest Order. Divide the L. C. M. of
the indices by each index in succession. Multiply the power and
root index of the first surd by the first quotient, of the second surd
by the second quotient, and so on.
Exercise 96.
Express as surds of the 12th order:
1. A^2; ^3; f^; 3^2; ^a^; ^1; i^^S.
^ 2. a/sS; 1^32; ^a^; v'^^X V^^^i"^.
Express as surds of the 7ith order, with positive expo
nents :
3. ^x^; V^; ah; ^'^j}; L; v/«~"; ^.
RADICAL EXPRESSIONS. 269
Reduce the following to equivalent surds of the same
lowest order:
4 V5, ^11, 4^; a/2, \^5, \/3; \^8, V3, ^6.
5. ^2, ^8, ^i; v^7, ^5, ^6; Va, ^a^
G. '^^, Va; ^^«, ^a6, ^a^ ^1^?, '^^^.
8. Vaic^, ^/a^Q^\ ^fm, "^n, v^, ^mnx.
9. v"^, \^6^ ^;?; 41^5^, 2^VlW^, 10 a a/37.
109. fV6=A/(IF^^=A/i =a/W.
i a/5 = a/()'' X 5 = V¥ = a/H .. IV'5>Ia/6
In general,
_ J
a J^x — (a" x)" = y^a* as,
5 ^'y = (6 y)* = .y^Fy. Hence,
I. To Compare Surds of the Same Order. Reduce them to
entire surds, and couipaie the resulting surd factors.
\ ^52"= ^{\y X 2« X 13 = y ^' =^^/42:25,
I ^8 = ^{\y X 2 = ^ (f)« X 22 = ^45.5625,
3 Vl = a/3* X f = '^3« X (fli» = >^46.656.
Therefore, the order of magnitude is 3 ^\, \ ^, \ >^52.
In general,
— ??J? JL
6 y^y = 6»»» ir ■ = "v^b^y*. Hence,
270 ELEMENTS OF ALGEBRA.
II. To Compare Surds of Different Orders. Reduce them to
entire surds of the same order, and compare the resulting surd factors.
Exercise 96.
Which is the greater ?
1. 3 V6 or 2 Vl4 ; 6 Vll or 5 VlSf ; 4 VG or 6 Vi
2. 10V5or4V3l; iVTorfVlO; ^^2 or ^3.
3. Vor^l; ^4 or ^5; Vf or ^T.
4. ^11 or '^f; 1.6 or J ylO; \^6^ or V^.
Arrange in order of magnitude :
5. V3, </4, </7 ; 8 V2, 5 a/5, 4 V7.
6. 2'^2l, 3^49, 4V7; 3^4, 4 ^I, 2^131.
7. 3 '^2, 3V2, A^4; 2^21, 3^8, 2V8.
Show that the following are similar surds :
8. ViO, V90, Vf ; J V'20, i V45, 5 Vf.
9. 7 V, 'V/ff , 3 VS ; ^162, 3 ^32, </2E92.
10. V27, Vr92, Vl47, Vl; a^W^^ h</W^^ f V— .
11 QiV*^ ^V5T2 ^K^l^ i/«^ .V«*^'^ ^.la^(?m^
RADICAL EXPRESSIONS. 271
110. Addition and Subtraction of Surds.
iVH"^=iV^x2 = fV2.
Adding,
3»a«Xa^arx(26)g 3 a o/——
26X726? =26^^^ ^ ^'
3/27a«x s/
°V 26 y2hxm = ~ 26 ^^^«'*'^'
1 8/46^ 13/4 62a: X a* 1 3/ — sttt
6V^ = ^V^=^3r^^ =^lJ/4a262x.
Adding,
= ^^y ^4a^b^x. Hence,
Reduce each surd to its simplest form. Prefix the sum or differ
ence of the rational factors to the common surd factor of the similar
surds. Connect dissimilar surds by their signs.
Exercise 97.
Simplify :
1. 3 Vis  2 \/20 + 3 V5 ; 3 \/ + 2 V^.
2. 2V + 3Vi; 2^l62J^^^; 3 V^ + ^?^.
272 ELEMENTS OF ALGEBRA.
3. A/3 + Vli2V5i; 5v^^=^542v/::r6 + i'^685.
5. S^J^+^MlW^; 3\/l627^!^32 + ^1250.
• 6. ^?+ 1^^ 3 V^27^2. ^403 v''320 + 4^^135.
7. V50ab^c^VS2aH^ (4:hc^3ac)V2al)c.
8. a; Vwi^^ 71^"^ x^ — m vm^ n^^ x^^ + n ^m'^ n^ aP.
9. V3 a Z^'^ + 6 a& + 3 a + a/3 a &2 _ (3 ^ ^ + 3 ^^
10. y/^+Y/^_2« + «^^V^^^.
^ a — ^ a { a^ — h^
11 ^A + 0SVfiVV96+1.5^ii^T750 + 8V.
111. Multiplication of Surds.
3 v/a X 7 V^ = 21 y4 X 3 = 42 y'3.
/^2 X ^3  lJ/2^"xT3 = :^432.
f V2 X 1^3 X lA^ X ^i = f X I X f ^£6 X 3^ X {\f X (i)«
= <{^28= ^2. In general,
a ^x X h y^y = a x"" X h y^ = ah {x ?/)« = a 6 y^ari/.
a yar X h 'J^y = ax"" X b y'^ = a &(»;'"?/")"'« =: a & y'a;"' i/**. Hence,
RADICAL EXPRESSIONS. 273
I. To Find the Product of two or more Monomials. Reduce
the surds to the same order (if necessary). Prefix the product of the
rational factois to the product of the surd factors.
Multiplicand,
Multiplier,
3V3,
2^6,
3V2
3V3H
9^6
9V«H
2>^5
h2^
3 >y/22v^5 multiplied by
3/y/22'v^5 multiplied by
Sum of partial products.
Hence,
6^5<^X3»
6 ^2«X 624/^3(7
h 6 v^288  6 ^'6754 v^30.
II. To Find the Product of two Polynomials. Proceed as in
Art. 24.
(^^/2\2^3){^^22^/3) = (3 X 2*)^(2 X 3*)^^ = 32 X 22^ X 3 = 6.
(a^x^b\/~y) (a^xbx/y) = (a x^^  (hy^f = a^xh^y. Hence,
III. The product of the sum and difference of two binomial
quadratic surds is a rational expression.
Exercise 98.
Simplify :
1. 2'v/r^ X 3 V3; SVf X J\/T62; J VlO x J^ Vl2j.
2\/l4 X V2i; 3^1 X gVJ.
(5 V35) X 2 V3; \!^64x2V2
4. (V2 + V3 + 2 \/5) X V2 ; 4 ^75 X 2 V^.
5. J V4 X v^iO; i VJ X § V^l; V5 X ^2.
2. J ^4 X 3 v^2
3. 3 >^3 X 3 V2
6. 3 \/ X ^1 ; J \/§ X 9 ^1 X \^.
18
274 ELEMENTS OF ALGEBRA.
7. 2^3 X'v/2 X J'^i; V^%X</'i; ^T68 x '^147.
8. '^^X^9X^9*; (3V23'v/6V8+3V'20)x3V2.
9. a/5 X V^IO ; (Vn  Vm) xVn; 4 \/^ x 3 VS.
10. Vmn X \^S m^x X V2 nx.
11. "^/m^no^ X "^m^ n x ; 2 Va X '^^ X 3 ^a x '^^.
12. ^^2^ X ^3^ X V i; ^^(4 7/^ a^'^)" X ^(:2m^x)\
13. VxtV^r^; (V23V3)(2V3 + 3V2).
71 ▼ 71 3 ▼ 2 ft*
14 (3V54^2)(2^5 + 3V2); (^2 + ^3)^.
15. (5 V3  6 a/2 + a/5) (lO a/3 + 12 a/2  2 a/5).
16. (V2 + </l+ '^D «/2  V3); '^24 X 6 ^3.
17. (V2 + V3) (V2  VS); (^3 + ^4) (^3  m
18. (a/5a/3)(V5 + a/3); (a/5 + 2 a/3) (a/5  2 a/3).
19. y^l2 + a/19 X y^l2 a/19; v'TG X a/S.
20. y'9 + A/n X \^9  A/rZ ; a/3 X ^2 X ^.
21. (^a3 + ^^) (^2 _ ^3) . y ^ ^1
RADICAL EXPRESSIONS. 275
22. V^10+ V68 X y^lO V68; v^^^xy^^.
23. {dVx+3 Va^x) (5 Vi  3 Va2^x),
24. ^S X ^M; VW X ^i; {mfi<^Mf'
112. To Rationalize Surd Denominators of Fractions.
2 2 X V3 2 V3 2
;^ = 7^^ 7 = ^ = T^ X 1.732 + = .23 + .
2 2X v/3^« 2 ^3« 2 ,/_
— =r= =  — !^, = —  — (n>m) = — T . Hence,
I. If the Fraction be of the Form — — = . Multiply both
terms by y"^^^^ ^ ^^
3+ V5 _ (3 + ys) X (3 + ys) _ 3« + 2 (3) ( /y/s) + (^5) *
3  V5 ~ (3  V5) X (3 + Vs) ~ 32  ( VS)^
14 + «V5 7 + 3X2.236+ ^ ^^
= 95 = 2 = ^•®^^"^
4V3 + 3a/5 (4 a/3 + 3^5) X (2V73 V2)
2 V7 + 3 >v/2 ~ (2 V7 + 3 V2) X (2 a/7  3 v^
_ 8 \/2l + 6 a/35  12^/6  9 a/To
g ox(A/^TA/g) _ aiVbTVc) _ a{\/bT\^c)
V^±A/^"(A/^iA/^)x(A/ftTA/^)" W~W~ ^"^
g a X Tfe T A/g) _ ajh^^c) _ a(hT Vd „
b±^~ (b±\^c)x(bTV~c)~ (pyic^^ " ^* ~ *^ ^"^'
276 ELEMENTS OF ALGEBRA.
II. If the Denominator is a Binomial Involving only
Quadratic Surds. Multiply both terms of the fraction by the
terras of the denominator with a different sign between them.
Note. It is often useful to change a fraction which has a surd in its de
nominator to an equivalent one with a surd in its numerator. Thus,
8 SXVI 8 I'S^ J X 2.236+= 1,3416+.
V5 V5 X V5 5
Exercise 99.
Eationalize the denominators of:
2 3 2  ^2 3 V5
1.
V2 + V3' 2 V5  V6 1 + V2 V3 + V2
85 V2 , 2 V"5  V2 1 6
3 2 a/2' V5 + 3V'2' 32 Vg' 'V^64 '
Vx — Vy , Sx — Vx y _ Va + a; + V<?' — x
o.
"s/x + Vy Va: y — 2>y Vet \ x — Va
X
. X — Vx^ — 1 a 1 2 a
4.
+ Vx^l' \/a+Vb V5V'2' 3a/2^^
Given V2 = 1.414, V3 = 1.732, V5 = 2.236 ; find
the approximate values of:
5. ^_; V50; 8K288 '' ' '
6.
V2 ^ V5 2V675 V500
1 + V2 1V5 3 1.1
2 + ^/2' 3+V5' 21/23^/3 ^5^2 2 + ^3
RADICAL EXPUESSIONS. 277
113. Division of Snrds.
2 V54 ^ 3 ^6 = I VV = f X 3 =2.
Ingeneral, a^i^6^y = ^^)" =^Vy*
I. If the Divisor is a Monomial Reduce the surds to the
same order (if necessary). Prefix the quotient of the rational factors
to the quotient of the surd factors.
, . , ,x 3\/3 3 V3X (3^32^2)
3 V3 r (3\/3 + 2 V2) = :^ p = 7 7= V\ / y A
^ ^ ^ 3'v/342V2 (3V3+2v/2)x(3V32V2)
^276V6 Hence, in general,
II. If the Divisor is a Binomial Involving only Quadratic
Snrds. Express the quotient in the form of a fraction, and ration
alize its denominator.
^a ^ \/b  \/c) a + 2 \/ab + bc {\^a+ \/b\^c.
Divisor multiplied by '\/a, a + \/a b — \/a c
First remainder, \/a 6 + 6 + V** c ~ c
Divisor multiplied by y'ft, \/a b + b — ^/bc
Second remainder, y'a c + ^b c — c
Divisor multiplied by \/cj ^ac + \/6 c — c
Hence, in genenil,
III. To Divide a Polynomial by a Polynomial. Proceed as
in Art. 33.
278
ELEMENTS OF ALGEBRA.
Exercise 100.
Simplify :
1. 21V384^8V98; 5 \/27 ^ 3 V24; \/l2^V^24.
2.  13 VT25 ^ 5 V65 ; 6 Vl4 H 2 ^21.
2V98 ' 7 a/22' 5V112 ' V394 ' ^2 * Vs'
4 IJ ^2  I Vll; ^12  ^2 ; V6  A^4.
5. 20 ^^200 4 4 a/2 ; ^18 ^ a/6 ; 4 ^32 ^ '^IG.
3 a/108 5 a/14 15 a/84
7. (15 a/105  36 v'lOO + 30 A^81) ^ 3 Vl5,
8. '^OOei^ViO; a^'a^c^^^; Va ^ \^.
m — 71 ^ m — n ^ {m ~ nf '3 »2
10. {acx^Vy—hcy ^/x) ^ c a/^ ; '\/a~x ^ ^'o^.
11. ^4 m 7^2 H V2w3^; v^2W^ X ^^^?^3^ aA;?^5
12. A^4^i2^XA^'9^^i2^*^v'25^^2^; <^d~^r\/^'
13. y— ^xV/2^— V/— • (ajl)f(A/ajl).
RADICAL EXPRESSIONS.
279
14. V10.4976 ^ 2 Vo ; (2 a:  Vo; y) ^ (2 Va: y  y).
15. (:^ a/3 + 2 V2) ^ (a/3 + V2) ; 4 ^a^ f 3 Vo^.
,^ 2A/T5 + 8 . 8V3+ 6a/5 84\/5 . 3a/57
Id. =:r 7 — —', 7^7 T pr •
5 a/15 5a/33a/5 1 + a/5 5 + a/7
17. (^x« + ^^v + w")  (V'^'^"  va;V + vy).
114. Involution and Evolution of Surds.
l^v/ir=[Mi)T=i.x®'=4v/I=^vi.
y486av/4a« = [3« X 20(220*)*]' = [3^ X 2^a^f = 3 X 2*0* = 3^/2^
m mp
In general, (a^i v^t^)^ = (a«i 6"*)' = a"!^ 6^ = a^h^ >v/6*^.
Vo'»i v^6™ = (a"*! 6" j^ = a »• 6" ^ Hence,
Express the surd factors with fractional exponents, and proceed as
in Art. 104.
Example 1.
V A«/;5 2a) ~ Ul 2aJ
©■3Gr(i).3@C4)'e)'
^ 'S)©"K)(^)
^?
2»a«
3a*
2cH
3q«
3
4a*c*
3
i«ra
8a»
8a«"
280
ELEMENTS OF ALGEBRA.
'i
CO
+
I
+
+
0
Li
i<5
^
CO
1
1
1 y
1
'>
1
^
+
+
"«
Ol
CM
II
Th
+
Hw
*
«
c
Tt^
^
I
HM
O
CO
1
^
%
GO
GO
+
+
Hn
c
c
^
Tf
'Ca
^5
a
r^
'd
S
S
c
^
1
fn''
fi"
.2
U
jT
pT
a;
J
^h"
'IS
ir^
M
c
s
s
^
(V
S
id
1
S
1
i
s
s
tJ
C
8
^
^
TIJ
n
s
02
CK
03
^
rt
a
G
rt
^
^
ti
^
£
•>(
o;
S
£
s
pR
pR
'/2
m
02
02
RADICAL EXPRESSIONS. 281
Exercise 101.
Find the values of the following :
1. mf; ^VE; i^Vlf; ^^2; (^32)'.
2. 'fe^; ^m, i'^I^jf; V^Gi.
3. {</uf; vQ^; mf; ^vff; mt
4 ^JV^; (^27)^ (^A; (2^3¥f ; 'sp^
5. (2 ^^6)'; y'V©"' '^"•^^■^"^ (2a^2Fo/.
6. ~^/•i•a'■, {Z'^Wc^f; ^iU^"; "v''27»=^.
7. 'v'»»v'^»; [\/(ac/]"; "^1; [jv^I^^^js,
11. v^9a:— ; [{x ^ y) V^y^  VnS^2i^^,
Find the values of the following, and express the results
in terms of positive exponents, by inspection :
12. (^^^T?f ; Wl + V\f; {V2 + Vsf.
282
ELEMENTS OF ALGEBRA.
13. (^3^l)MV^+^i) ["^fl
15. M^.o]^ [(^«f +(fr^)T
16.
[jn^ J'Zm 2^m T ["^ /m^ 4 ^^3a "[
12 mV n "^ 4/n J ' L V ^4 ^^^^^3 J
Find the square roots of :
17. ^ + 1 + 9 ^^ + 1^1 _ 1^1 ^^ _ 6 ^^.
18. 1  2"+^ + 4"; 9" — 2"+^ X 3" + 4".
19.
r
V2.
ic 4?/ ^a; '2/
m
20. v"^  4''^2;5 + 4 :i:"^ + 2 a/^^ "'  4 'v^^^ '" + a/^'
Miscellaneous Exercise 102.
Find the values of the following :
1. 'V^; v^; V3xV27; 1""; "'^4; a/X^V.
2.' ^^32^; V^e^m^^ V^m ^ V^
Reduce the following to their simplest forms :
18' ^'2'
3. t J^,; V^; ^3888; ^!^±^sj^
»)
RADICAL EXPRESSIONS. 283
Reduce to equivalent surds of the same lowest order :
5. V2, '^, •^5; v^y, a!^8, 4/5; 3^75, 3v^54.
6. 2A/I8, ?,\/U, 4A!^r62, 5^128; V^\ </V, i^^.
Change to similar surds :
7. •^27, ^144; C^Si, 3^; W2, ■^243.
8. ^02, VI; VTo, 4'Wb; I'^lh, V^.
9. 2\/^. ^726^, ly^^; K.1G; 4^275.
10. n, VJl, ^i^, 3 ^^; V20, 3 '^. 4 Vl25.
11. ^32. ^128; VM^. y/^'. SfW^'
12. ^J^. ^192, ^S". ^^. ^A; ^2, ^2?.
Arrange in order of magnitude :
13. 3 a/2^, 4 </^, 3 ^3 ; 5 ^8, 3 ^9, 3 VlO.
14 fnfV3~, iV5, 2^; Vi ^H
15. a' e^^sTjy, ^ ^(25^^, (64)" sj"^, .
Find the values of:
16. V243 4 \/27 4 v'48; 2 V^189 + S'l^'STS 7^,
284 ELEMENTS OF ALGEBRA.
17. 4 V5 X 'V^lT; 3 'v'^gOO ^ \/5; \J2 V 2Vl ^ ^V^^'
18. 5V'2 + 3V82V32; 3 ^^ST  4 ^1^192 + ^648.
19. 'V^m X #432 ; I V5 X  #2 x #80 x #5.
20. ^64+5^32^^108; i(^27 +  V^i92 + •^81).
21. 2'v^JJ^ + a/60  \/225  Vf ; J V^ { a/2 + 3 V )•
22. (a^^2^^)V^I6; (^92^21 + 4^1)2^9.
23. (6 ^1 + ^18) f ^72 ; 1 ^/20  3 v^5  v/f
24. J,a ^60^(2 ^240 + 7^31); y/j^^^^)*.
25. (^ro 2^4 + 4 ^54) (o ^64 + sV^ 2 ^32).
Eationalize tffe denominators of:
^g V2Q  a/8 . (3+ V3)(3 4 #5) (#5  2),
* V5 + V2 ' (5  V5) (1 + VS)
^ m \ (171 — Vvi^ + a? 71^ x"' 2
m — a ?l + V??i2 ^ ^2 7^2 „ ' ^5 __ /y/3 _ ^2
Simplify the following :
28. ^_ ; V(fy" X V(f ; #(8'a3&)2 x #(2 a &3)2.
V2
^^ 7 + 3 \/5 _^_ 73a/5 ^ /^9?»Ti x V3l<T"
^" 73^5 7 + 3A/5' V 3V3^
30
. {a r #«)" + ' y^(^ j/^"")"'; #^2^^""* X \x^yl
RADICAL EXPRESSIONS. 285
31. 4^aX'i^:r^X ^a* X "/a X ^aV x ^^.
33. 4.x^<77xf x^^^„^(^y^^/*^
34. V(a>) ^ v/(^.,p; 1^,; ()t.
35. .^ (g) ^ Vl^. ; I ^1 + 1 ^^'  2 (S)«.
37 V^ + «^ V^ ^^ ^ 4 ^^
38. ^Hl^^^i H f 1 + ^V; i^a^ + ^a)3.
^5 X ^^3
40. Express r^ with a single radical sign.
Queries. What sign is given to the Titli ])ower ? To the nth
root? Why? How change the order of a suixl ? In T., Art. 112,
why take m less than n ? How rationalize a sunl denominator ? What
powers of n^ative nuniWrs are positive ? What n^ative ?
286
ELEMENTS OF ALGEBRA,
Imaginary Expressions.
115. An Imaginary Expression is an indicated even root
of a negative expression ; as, V— a ; a \h V— 1. V— 1
is an imaginary square root ; a V— 1 is an imaginary
fourth root; etc.
V^T^^ :^ ^a2 X ( 1) = V«^ X V^ = « a/^^
/^TJ  .^6 X ( 1)  ^b X V^ Hence,
Every imaginary square root can he expressed as the product of a
rational or surd factor multiplied by \/— 1.
The successive powers of ^y/— 1 are found as follows .
)ip=(l)i = + V=T;
)*/=(!) =l;
)*r = (i)' = (i)(i)* = V~;
)J]'=(l)^ = + l;
)i]«=(.l)» = l;
)i]»=(l)^ = + l;
)J]» = ( 1)1 = ( i)^( i)i = + ^—; and so
)*]'=( 1)5 =VPT)= ±v/^orTl,
oc?c? or eyen integer. Hence,
The successive powers of \/~ 1 form the repeating series :
+ V~h h V^» +1
The methods for operating with imaginary expressions are the
same as those for surds ; but before applying the methods it is better
to remove the factor /y/— 1. All cases of multiplication can be made
a direct application of Arts. 97, 114.
w
1? =
= [(1
w
xY
= [(1
w
~xY.
= [(1
w
lY
= [(1
w
^Y
= [(1
[V
xY
= [(1
[V
1]' =
 [(1
w
T]' =
= [(1
w
T/ =
= L(i
on. In
general,
W~
T]" =
=[(1
accordin
gas
n is an
RADICAL EXPRESSIONS. 287
niustratioiiB. ^/ 6aH^= y/S a^b* X (1) = \/S aH^ X \/l
= 2ab \/Tb X V^.
V9«^ + V49a«V4a'«= 3a ^ 1 +7a/v/ 1 2 a
= 10a V^ 2a
= 2 a (5 V=n  1).
3 >v/^ X 4 v^^ = (3 V^ X \/^)("* V^ X V^)
= 3 ^3 X 4 y 2 X ^/^ X V^
= 12V3X^X [(l)*]'
=  12 ye.
2 v/^ X 5/v/^ X 3 V^
= 2 V3 X 5 V2 X 3 >v/6 X \/^ X ^^^ X V"^
= 30>v/3 X 2 X 6 X [( l)*]"
=  180 ^~l.
= f v/3 X 1 = J V^
_ (i + yH)' _ i+.2v^_+ (0
Example. Multiply 12 V"^ by 3 + ^y/^.
ProcesB. 1—2 ^—~i
3+ V1^
1  2 y/ 1 multiplied by 3, 36 V" ^
12 y'lH! multiplied by y'^, 2 + V^
Sum of the partial products, 6 — 5 /y/— I
288 ELEMENTS OP ALGEBRA.
Notes : 1. Imaginary expressions represent impossible'operations ; yet it is
a mistalie to suppose that they are unreal, or that they have no importance.
2, If the student employ the method of multiplying or dividing the expres
sions under the radicals (Arts. Ill, 113), for all cases in multiplication and divi
sion, he cannot readily determine the sign of the product or dividend. Thus,
V^^ xV—a = V—aX—a= Va^ = ±a.
3. Is the above product both ±a or — a ? We are limited to the considera
tion of the product of two equal factors, and we know that the sign of each is
negative ; also, that Va^ = it «. Hence, the sign of Va^ will necessarily be
the same as that of each of these factors. Therefore, it will be the same as was
its root. Thus,
V 3 X 1/ 3 =  1/9 =  3,
Exercise 103.
Simplify :
1. V^; '^16; V 12 a; V^^T^; V^
2. V49a2&6. ^^7729; ^IT^; y'^^^".
Find the values of :
3. (V— i)i^,(V^f ; iVif; {V—lf.
4. iV^lf; (V=^r; iV=lf; {V^f
5. A/25  A/49 + V121  a/64 + V1 a/36.
V22 V216
6. 2 V 24 + —=  V 18 ; ^...^ 
V3 A/33 V324
7. V 36 a^ 4 V 9 a^  V (1  af a^  V a\
8. V{cthf+ V(a2 2ab + b^)+ V1 6 a^ b^V 4 a^
Multiply :
9. V^ by V^; 3 V^ + V^^ by 4 V^^
RADICAL EXPRESSIONS. 289
10. 2 V^ by 4 V'^; 1 + V^ by 1 + V^.
11. V 2 + 3 V^ by V^^ + 3 V^.
12. 32 V4 by 5 + 3 V^^; 4 + V^ by 4 V^.
13. 1 + V^ by 1 V1; 2  V=^ by 1  2 V^~3.
14. 2 V^  6 V^ by V^ + V^.
15. Va — ^ by V^ — a ; a + V— a; by a — V— a;.
16. a V— a + b V— b by a V— a — 5 V— 6.
Divide :
17. V^^ by V^^;  \/^ by  6 V^.
18. V^ by V 20; V 24  V^ by a/^^
19. 2 V— 4 «*» by V— a^ \ a + V a by V a^.
20.  2 V^ by 1  \/^ ; 2 by 1 + V^.
21. \^^^^ by v^ 5; '^^^^ by ^=^.
22. 4 + V^ l)y 2  V^; V^3 by 1  V^.
Rationalize the denominators of :
23 ^i^J^^^ 2 \/:ri _ 3 yry 3 + 3 V^
" ' 2  V^' 4 a/^ + 5 V^^' 22 V=I '
Queries. To what form can all imaginary monomials be reduced ?
In multiplication and divi.sion why separate the imaginary expres
sions into their sunl and imaginary factors ? Is it necessary in all
?
19
290 ELEMENTS OF ALGEBRA.
Quadratic Surds.
116. I. A quadratic surd cannot equal the sum or differ
ence of a rational expression and a quadratic surd.
Proof. If possible, let ^/a = 6 db ^\fc, in which ^a and y^c
axe dissimilar quadratic surds, and 6 a rational expression.
Square both members, a = 6"^ ± 2 & ^/c + c.
± a T ^^ T c
Transpose, ± 2 6 \/c = a — ft^ — c*. .♦. ^/c
26
That is, a surd equal to a rational expression, which is impossible.
Therefore, ^\/a cannot equal h ± ^ c.
II. i/" a + Vb = X + Vy, in which a and x are rational
and Vb and Vy cire quadratic surds, prove that a = x and
b = y.
Proof. Transposing, /y/6 = (x — a) + V^?/ Now if a and a; were
unequal, we would have a quadratic surd equal to the sum of a ra
tional expression and a quadratic surd, which, by L, is impossible.
Hence, a = x. Therefore, ^Jh = ^^y, ot b = y.
III. 7/" V a + Vb = Vx + Vy, prove that y a — Vb
= Vx — vV
Proof. Square both members, a + \/b = x + 2 aJx y + y.
Therefore IL, a = x + y {!) and a/6 = 2 ^/x^ (2)
Subtract (2) from (1), a  \/b = x ~ 2 \/xy + y.
Extract the square root, V a — \/b = y\/x — \/y.
Similarly it may be shown that if V « — ^/b = ^/x — /y/jr,
then V a + ^Jb = ^Jx + ^/y.
RADICAL EXPRESSIONS. 291
Square Root of a Quadratic Surd.
117. To find the square root of a binomial surd a ± Vh.
Process. Let Va ± a/6 = V^ ± Vi^ (1)
Then (III., Art. 116), Va T V^ = V^ T \/y (2)
Multiply (1) and (2) together, ^af^ b = x y (3)
Square (1), a ± aA = x ±2 ^x y + y.
Therefore (II., Art. 116), a=xhy (4)
Add (3) and (4), a + v^^"^ = 2 x. .. x = ^
, a  Va* — b
Subtract (3) from (4), a  V a* b = 2y. .'. y = ^
Therefore, V7±Vb = \J ^^ V^^ ± \/" ~ ^"'~^ (0
HotM; 1. Evidently, unless d^ — h be a perfect square, the values of Vx
and Vy will be com plex surds ; and the expression Vx + \ y will not be as
simple as V a + 1/6.
2. Since, Va'^c + Vbc — \/c{a ^ Vh\ also if a^  6 be a perfect square
the squats root of a + Vh may be expressed in the form Vx 4 Vy, the square
root of Vcflc ■{■ Vbc'is of the form \'c ( V'x f Vy ).
3. Frequently the square root of a binomial surd may be found by in
si>ection. Thus,
FiTid ttoo numbers whose sum is the rational term^ and whose product is the
square of half the radical term. Connect the square roots of these numbers by
the sign of the radical term.
Examples : 1. Find the square root of 3^  VlO.
Process. Let Vi  Vy = "^H  aAo (1)
Then (III., Art. 116), y'i f Vy = V^T'^ (2)
Multiply (1) and (2) together, xy= \/*^ _ lo =  (3)
Square (1), x  2 V^ 4 y = 3i   /y/lO^
Therefore (II., Art. 116), z + y = 3 (4)
From (.3) and (4), x = 2, y=l.
Therefore, ^3^  \/Tb = a/ I = ^ V^ 1
292 ELEMENTS OF ALGEBRA.
We may employ the general form (i). Thus (since a = 83 and
y6 = + 12 V35),
L^ ^ ^g^2 _ (2 ^35)2
2. V83 + 12 ^/3b = \ ^^ — 
4/ 83  V8'3^  (12 V35f _ / 83 + 43 / 83  43
= ^63 + ^20 = 3 v^ + 2 Vs.
3. Va/ST  2 a/6 = V V3 (3  2 a/2) rr: ^3 X a/3  2 a/2, ^
also \/3  2 a/2 (in which a = 3 and a/& =  2 a/2)
^ i / 3 + V3^  (2 A/2y _ i / 3  V3"^ (2 a/2)^ ^ ^ _ i
.. a/ a/27  2 Ve = a!^3 (a/2  l).
4. Find by inspection the square root of 103 — 12 a/h.
Solution. The two numbers whose sum is 103 and whose
product is (^ — — j , are 99 and 4. Hence, Vl03  12 \/Tl
=: a/99  a/4 = 3 a/iT  2.
5. Similarly, VlO + 2 a/21 := a/7 + Vs, because 7 an^ 3 are
the only numbers whose sum is 10 and whose product is (a/21) • ^ ^
Exercise 104.
Find the square roots of:
1. 72A/rO; 5 + 2V6; 41241/2; 2J + V^.
2. 188 V5 ; 11 + 2 VSO ; 13  2 a/42.
3. 15~V56; 47 4 a/33; 62^5; 10 + 4 a/6.
RADICAL EXPRESSIONS. 293
5. V27 + a/15 ; 2?7i + 1 + 2 Vm^ + 7i  2.
6. (m^ + m) ?i  2 vi n Vm ; 9 — 2 VU.
7. (wi + w)'^ — 4 (m  /O A/m?i ; 3 a;  2 a: V2.
Find the fourth roots of:
8. 9756V3;  a/5 + 3J ; 56 + 24 V5.
9. 17 + 12V2; 4(31 8 Via); 248 + 32 V^.
Simple Equations Containing Surds.
118. Examples : 1. Solve V4 ar^  7 x + 1 = 2 a;  4 (1)
Process. Square (1), 4x^7x+l = 4x^7\x + ^.
.'. X = ]l\. Hence,
To Solve an Equation containing a Single Surd. Arrange
the terina bo as to have the surd alone in one member, and then raise
each member to the power indicated by the root index.
Note. If the equation contains two or more surds, two or more operations
may be necessary in order to clear it of radicals. Thus,
2. Solve \/'2b x1% ^4x 11=3 ^/i.
Process. Transpose, \/25a:29 = 3 ^/x 4 *J\x\\ (1)
Square (1), 25x29 = 9x+6 V(4a: ll)a:+4a:ll.
Transpose, etc., 'v/(4a; 11) x = 2 x — 3 (ii)
Square(2), 4x« llz = 4a;«12ar + 9. .. x = 9.
294 ELEMENTS OF ALGEBRA.
3. Solve ^^7= = ^ (I)
\ X + n ^x + 3n ^ '
Process. Clear (1) of fractions, transpose and unite, etc.,
/ ^ r TT / mn ( mn \2
(m — n) \/x = mn. Hence, \/x = . .*. x =
^ . \/m + X 4 ym — x
4. feolve ^' , J — n.
\m \ X — 's/m — X
Process. Rationalize the denominator,
fn. + 's/m'^ — x''
(1)
From (1), y'rn^  x"^  nx  m (2)
Square (2), m^ ~ x^ = n^ x^ "Imnx ■\ m\
Transpose, etc., a;^ (l + n^) =r 2 mna:.
Divide bv a;, a:(l + w^) =: 2mn. .*. a: i= 7— — ;,
Exercise 106.
Solve :
1. V^ + 5 = 4 ; V3 ^^ + 6 = 6 ; ^x^2^x^%
2. Va:2 _ 3 ^. _}_ 5 ^ :i; _ 1 ; ^2^3 1=2.
3. V'3 + V4 + V^^^= 2; Vr+W^^m = a: + 2.
4. ^/mx^~a = 'v/c^TT ; V^rr2 + ^4  ^3^ = 2:.
5. VJT~2 =. 2 V'2ic3 ; V3 a: + 5 = 3 ^"Ix ^ 1.
6. V3.r 4= V2a: + 16; ^^2 ^'  4 = V 4  V2 rr.
7. 3Vi =
RADICAL EXPRESSIONS.
8
295
V9 a;  32
4 V 9 a;  32.
8. Va: + 3 4 Vx + S  V4 :c + 21 = 0.
9. ^Vx+  ^Vx — 5 = y/2 Vi.
Q
10. 'V^m^ f a; Vn^ + x^ = Vx{ vi; Vx + Vx—2 = —r'
\ X
11. A^4+2V2^5 = V3; V + :^^m_
V a: — V m ^
V2 ic + 1 + 3 Va; _ Vm a; — n __ 3 v m a; — 2 7i
a/2 a; + 1 — 3 Va; Vwi x\ n 3 Vw x + 5n
13
V5a;+ VB Va;4 5 V6a;+2 _ 4:V6x + 6
V3^ 4 V3 ~ V^ + 3' V6^ 2 ~ 4 V6'^  9
14. v/^+v/^^=c/4^
"w + a; 'm — a; »7?r — ar
Solve the following for x and y :
15. a; + 4 V3 + y = 15 — a; 4 y V5.
16. a; 4 y 4 a; Va 4 y V^ = 1 — Va.
17. a:  5 + (2 y  3) V3 = 5 a;  Vl2.
18. a; — rt 4 {y — 3) Va 4 2> = w a; 4 Vol.
19. a — Va; + y = y — x — Vm 4 71.
20. a; Vw(Vm4 1) = w — wi4 y V^(l — V^.
296 ELEMENTS OF ALGEBRA.
CHAPTEK XX.
LOGARITHMS.
119. If a^ = m, then / is called the logarithm of m to the
base a. Hence,
A Logarithm is the exponent by which a certain num
ber, called the base, must be affected in order to produce a
given number.
The logarithm of m to the base a is written logam. Thus,
log„m = I expresses the relation a^ = m; logj^ 100 = 2 expresses the
relation 10^ = 100, etc.
Since numbers are formed by combinations of tens, any number
may be expressed, exactly or approximately, as a power of 10. Thus,
1000 = 103 . eic
120. Common System of Logarithms. This system has
10 for its base, and is the only one used for practical
calculations. Thus,
Since 100=1, log 1 = 0; since 10^ = 10, log 10 = 1 ;
since 102 = iqO, log 100 = 2 ; since lO^ = 1000, log 1000 =: 3;
since 10* = 10000, log 10000 = 4 ; and so on.
Since lOi = J^ = .1, log .1 =  1 = 9  10 ;
since 10 2 = ^i^ = .01, log .01 =  2 = 8  10 ;
since lOs = ^Jq^ = .001, log .001 =  3 = 7  10 ; and so on.
It is evident that the logarithm of all numbers greater than 1 is
positive J and of all numbers between and 1 is negative ; also, that
the logarithm of any numbers between
LOGARITHMS. 297
1 and W is f a fraction;
10 and 100 is 1 + a fraction ;
100 aiid 1000 is 2 + a fraction ;
1 and .1 is — 1 f a fraction, or 9 + a fraction — 10;
.1 and .01 is — 2 + a fraction, or 8 + a fraction — 10;
.01 and .001 is — 3 + a fraction, or 7 + a fraction — 10; and so on.
It thus appears that the logarithm of a number consists
of an integral part, called the characteristic, and a I'ractional
part, called the mantissa.
The mantissa is always made positive.
IUu8tration8. It is known that log 5 = 0.69897; log 12 = 1.07918;
log 2912 = 3.4(3419; etc. These results mean that loo«»8»' = 5;
1O1.07918 ^ 12; 108.4M19 = £912 ; CtC.
Notes : 1 . 'Hie fractional part of a logarithm cannot be expressed exactly,
but an apiiroxiniate value may be found, true to as many decimal places as
desire^l. Thus, the logarithm of 3 is found to be 0.477121, true to the sixth
place.
2. For brevity the expression "logarithm of 3" is written log 3. The
expression "log «" is read "logarithm of x."
3. Logarithms were inventetl by John Napier, Baron of Merchiston, Scot
land, and first published in 1614.
4. Ihre are only two systems of logarithms in general use : the Natural,
or Hyperbolic, system, and the lirhjtjsian, or Common, system. The base sub
script of the former is e, and that of the latter is 10.
5. The natunil system, invente<l by Jolm Speidell and published in 1619, is
employetl in the higher branches of analysis and in scientific investigations ;
its base is 2.718281828+ .
6. The common system, more properly called the denary or decimal sys
tem, was invente<i by Henry Brigps, an Engli.sh geometrician, and first pub
lished in 1617. The logarithm of its base, 10, is alrmys 1.
7. The logarithms invente<l by Napier are entirely different from those in
ventetl by Si>eidell, though they are closely connected with them. The natural
system may be regardeil as a modification of the original Napierian system.
121. Since log 1 = 0, log 10 = 1, log 100 = 2, log 1000 = 3, etc.,
the characteristic of the logarithms of all numbers consisting of one
298 ELEMENTS OF ALGEBRA.
integral digit (that is, all numbers with one figure to the left of its
decimal point) is 0; of all numbers consisting of two integral digits
is 1 ; of all numbers consisting of three integral digits is 2 ; and so
on. Hence,
I. The characteristic of the logarithm of an integral
number, or of a mixed decimal, is one less than the numher
of integral places.
Since log .1 =  1, log .01 =  2, log .001 =  3, etc. ; the charac
teristic of the logarithm of any decimal whose first significant figure
occupies the first decimal place (that is, of any number between 0.1
and 1) is — 1 ; of any decimal whose first significant figure occupies
the second decimal place (that is, of any number between 0.01 and
0.1) is — 2 ; of any decimal Vfho^Q first significant figure occupies the
third decimal place (that is, of any number between 0.001 and 0.01)
is — 3; and so on. Hence,
II. The characteristic of the loga7nthm of a decimal is
negative, and is numerically equal to the numher of the place
occupied hy the first aignificant figure of the decimal.
The characteristic only is negative. Hence, in the case of decimals
whose logarithms are negative, the logarithm is made to consist of a
negative characteristic and a positive mantissa. To indicate this, the
minus sign is written over the characteristic, or else 10 is added to
the characteristic and the subtraction of 10 from the logarithm is
indicated.
Thus, log .0012 = 3.0792, or 7.0792  10 ; read "characteristic
minus three, mantissa nought seven ninetytwo," or "characteristic
seven minus ten, etc." In reading the mantissa, for brevity, two inte»
gers are read at a time. Thus, log 2 = 0.30103, is read "the loga
rithm of two equals characteristic zero, mantissa thirty ten three."
Illustrations. The characteristic of the logarithm of 9 is 0;
of 32 is 1 ; of 433 is 2 ; of 39562 is 4; of 632.526 is 2 ; of .42 is  1 ;
of .023622 is  2 ; of .0000325 is  5 ; etc.
LOGARITHMS. 299
122. Let m and n be any two numbers whose logarithms are x
and y in the common system. Then l(F = in and 10*= n. Multiply
ing the equations together, we have 1(F+*' = mn. Hence (Art. 1 19),
log mn = z + y. But x = log m and y = log n. Therefore, log m n
= log m H log n. Similarly log in np = log m f log n + log p ; etc.
Hence,
Tlie logarithm of a product is found by adding together
Die logarithms of its factors.
Ulustrationa. Given l(»g 2 = 0.3010, log 3 = 0.4771, log 5
= 0.6990, log 7 = 0.8451.
log 252 = log (2 X2X3X3X7)
= log 2 + log 2 + log 3 f log 3 + log 7
= 2 log 2 + 2 log 3 + log 7
= 2 X 0.3010 + 2 X 0.4771 + 0.8451
= 0.6020 + 0.9542 + 0.8451
= 2.4013.
log 300 = log (2 X 3 X 5 X 10)
= log 2 + log 3 + log 5 + log 10
= 0.3010 + 0.4771 + 0.6990 + 1
t= 2.4771.
Exercise 106.
Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990,
log 7 = 0.8451 ; find the values of the following :
1. log 6; log 64; log 14; log 8; log 12; log 15; log 84.
2. log 343; log 16; log 216; log 27; log 45; log 36.
3. log 90; log 210; log 3600; log 1120; log 1680.
123. If any number be multiplied or divided by any integral
power of 10, since the sequence of the digits in the resulting number
remains the same^ the mantissa) of their logarithms will be unaffected.
Thus, since it is known that log 577.932 = 2.7619,
300 ELEMENTS OF ALGEBRA.
log 5779.32 = log (577.932 X 10) = log 577.932 + log 10
2.7619 + 1 = 3.7619.
log 57793.2 = log (577.932 X 100) = log 577.932 + log 100
2.7619 + 2 = 4.7619.
log 57.7932 = log (577.932 X 0.1) = log 577.932 + log 0.1
2.7619 + ( 1) rr 1.7619.
log 5.77932 = log (577.932 X 0.01) = log 577.932 + log 0.01
2.7619+ (2) = 0.7619.
log .577932 = log (577.932 X 0.001) = log 577.932 + log 0.001
= 2.7619 + (3) =1.7619.
Etc. Hence,
The mantissce of the logarithms of numbers having the
same sequence of digits are the same.
Illustrations. If log 44.068 = 1.6441, log 4.4068 = 0.6441,
log .44068 = 1.6441 or 9.6441  10, log .000044068 = 5.6441 or
5.6441  10, log 440.68 = 2.6441, log 440^800 = 6.6441, etc. If
log 2 = 0.3010, log .2 = 1.3010, log .02 ^ 2.3010, log 20 = 1.3010,
etc. Hence,
The mantissa depends only on the sequence of digits, and
the characteristic on the position of the decimal point.
Exercise 107.
1. Write the characteristics of the logarithms of : 12753;
13.2; 532; .053; .2; .37; .00578; .000000735; 1.23041.
2. The mantissa of log 6732 is .8281, write the logarithms
of: 6.732; 673.2; 67.32; .6732; .006732; .000006732. .
3. Name the number of digits in the integral part of the
numbers whose logarithms are: 5.3010; 0.6990; 3.4771.
LOGARITHMS. 301
4. Name the place occupied by the first significant fig
ure iu the numbers whose logarithms are: 4.8451; 0.7782.
Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.G990,
log 7 = 0.8451; find the logarithms of the following
number :
5. .18; 22.5; 1.05; 3.75; 10.5; 6.3; .0125; 420.
6. .0056; .128; 14.4; 1.25; 12.5; .05; .0000315.
7. .3024; 5.4; .006; .0021; 3.5; .00035; 4.48.
124. Let 7« be any number whose logarithm is x. Then UF^wi.
Raising both members to the pih power, we have 10'" = inP. Hence
(Art. 119), log mf = px. But x — log?n. Therefore, log Tn**
= p log m. Similarly log wi^'n' = p log m \ q log n, etc. Hence,
Tlie logarithm of any power of a number is found h/
multiplying the logarithm of tlie number by the exponent of
Vie power.
niustrations. log 5" = 10 log 5 =J0 X 0.6990 = 6.99(H),
log .003* = 5 log .(K)3 = 5 X 3.4771 = 13.3855.
log 864 = log 26 X 3« = 5 log 2 + 3 log 3 = 5 X 0.3010 + 3X0 4771
= 1.5060.
Note. If the number i.s a decimal and the exponent positive, the j)roduct of
the characteristic and exponent will be negative, and since the mantissa is made
positive, we must algebraically add whatever is carried from the niantis.sa.
Thii.«», log .0005" = 10 X 4.6990 = 40 + G.9900 = 34 + 0.9900 = 34.9900.
Exercise 108.
Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990,
log 7 = 0.8451 ; find the logarithms of:
1. 2*; 53; 7^; 8^ 3^; 64; 81; 72; (8.1)7; (2.10)6
2. 343; .036; .000128; (.0336)1^ (.00174)2; (3.84)»
302 ELEMENTS OF ALGEBRA.
125. Let m and n be any two numbers whose logarithms are x
andy. Then 10^ = m and 10^' = w. .. 10^2' = m f n.
m " mn
log  = a:  2/ = log m  log n. Similarly, log ^^^ = log m + log n
— (log wij + log Uj). Etc. Hence,
The logarithm of a quotient is found by subtracting the
logarithm of the divisor from the logarithm of the dividend.
Illustrations, log  = log 3 log 2 = 0.47710.3010 = 0.1761.
log f = log 5  log 7 = (0.6990)  0.8451 = (1.6990  1)  0.8451
= 0.8539  1 = 1.8539.
Note. To subtract a greater logarithm from a less logarithm. Add to the
characteristic of the minuend the least number which will make the minuend
greater than the subtrahend ; also indicate the subtraction of the same number
from the minuend so increased. Then proceed as before. Thus,
log =^ = log 252  log 300 = (2.4014)  2. 4771 = (3.40141)  2. 4771 = 1.9243.
log '^ = (3.6990)  2T8451 = (T.6990  2)  2.8451 = 0.85392 = 2.8539 or
8.8539  10.
Exercise 109.
Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990,
log 7 = 0.8451 ; find the logaritlims of:
1 ^
2'
_3^ 7 3 .003 .005 /SV .007
' .05' 5' 5' 2' Q7 ; 02 ' VlO/ ' 02'
2 42. :!!_. 125 '^ ^ ^i^. 5. ^
■ ^' .0052' ' 5 ' 8.1' .000027' ' .007*
126. Let m be any number of which the logarithm is x. Then
X
1(F = m. Taking the rth root of each member, we have 10'" = \/m.
.'. (Art. 119), log \/m —  = ^ — . Similarly, log ^^m n
LOGARITHMS. 303
The logarithm of any root of a numher is found hy divid
ing the logarithm of the number by tJie index of the root.
«/ log 5 0.6990 ^ _„
IHuatrations. log ^ =  = — 5— = 0.1398.
,, log.0(X)7 4.8451 3.8451 + 7 ^„  ,,^«
log '^/ioOO? = ^ = y— = 7^ = 0.5493+1 = 1.5493.
1 VT^. logl5<^ 5 log 3 X .5 5 (log 3 + log .5)
logVl5»=— ^r— = e = 6
^5(0.4771 + 1.6990)^^^^^^^
o
Note. If a negative characteristic is not exactly divisilile by tlie index of
the root, subtract from the characteristic the least positive number which will
make it so divisible. Indicate the addition of the characteristic so formed to
the mantissa, and prefix the number subtracted from the characteristic to the
mantissa. Then divide separately. Thus,
log V75 = ?5:5 = I:^ = '•6»y + ^ = 0.8495 + T = T.8495 or 9.8495  10.
Ik ii it
Exercise 110.
Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990,
log 7 = 0.8451 ; find the logarithms of:
1. ^7; \^l; \/2 v^.^; ^243; ^12^; \^M; ^^.
2. ^^iTK2f; 5ix3i; ^,; g; ^^^; 6i x 3f.
,, ^^x\/2 j^^ ^ 7/ J.21)2_ V^2^
"• ;/l8xV2^ ^15' ^!^7' V(.()()084)2' ^^^^^fg '
127. Table of Logarithms. The table (pages 304 and 305)
gives the nianti.Hsae of the logarithms to four decimal i>]jices for all
numbers from 1 to 10(X) inclusire. The characteristic and decimal
points are omiUedy and must be supplied by inspection (Art. 121.
304
ELEMENTS OF ALGEBRA.
N
1
2
3
4 6
6
7
8
9
10
11
12
13
14
ouoo
0414
0792
1139
1461
0043
0453
0828
1173
1492
0086
0492
0864
1206
1523
0128
0581
0899
1239
1558
0170
0569
0934
1271
1584
0212
0607
0969
1803
1614
0253
0645
1004
1335
1644
0294
0682
1038
1367
1673
0384
0719
1072
1899
1703
0874
0755
1106
1480
1732
15
16
17
18
19
1761
2011
2304
2553
2788
1790
2068
2330
2577
2810
1818
2095
2355
2(501
2833
1847
2122
2380
2625
2856
1875
2148
2405
2648
2878
1908
2175
2430
2672
2900
1931
2201
2455
2695
2928
1959
2227
2480
2718
2945
1987
2253
2504
2742
2967
2014
2279
2529
2765
2989
20
21
22
23
24
3010
8222
8424
3617
3802
;;o32
3243
3444
3636
3820
3054
8263
8464
3655
3838
3075
3284
3488
3674
3856
3096
3304
8502
3692
3874
3118
3324
3522
8711
3892
3139
3845
8541
3729
3909
3160
3365
3560
3747
3927
3181
3385
3579
3766
3945
8201
3404
3598
3784
3962
25
26
27
28
29
3979
4150
4814
4472
4624
8997
4166
4330
4487
4639
4014
4183
4346
4502
4654
4031
4200
4362
4518
4669
4048
4216
4378
4533
4683
4065
4282
4393
4548
4698
4082
4249
4409
4564
4713
4857
4997
5132
5263
5891
4099
4265
4425
4579
4728
4116
4281
4440
4594
4742
4133
4298
4456
4609
4757
30
31
32
33
34
4771
4914
5051
5185
5315
4786
4928
5065
5198
5328
4800
4942
5079
5211
5340
4814
4955
5092
5224
5353
4829
4969
5105
5237
5866
4843
4983
5119
5250
5378
4871
5011
5145
5276
5403
4886
5024
5159
5289
5416
4900
5038
5172
5302
5428
35
36
37
38
39
5441
5568
5682
£798
5911
5453
5575
5694
5809
5922
5465
5587
5705
5821
5933
5478
5599
5717
5882
5944
5490
5611
5729
5848
5955
5502
5623
5740
5855
5966
5514
5635
5752
5866
5977
5527
5647
5763
5877
5988
5539
5658
5775
5888
5999
5551
5670
5786
5899
6010
40
41
42
43
44
6021
6128
6282
6335
6435
6031
0138
6243
6345
6444
6042
6149
6253
6355
6454
6053
6160
6263
6365
6464
6064
6170
6274
6375
6474
6075
6180
6284
6385
6484
6085
6191
6294
6395
6493
6096
6201
6304
6405
6508
6107
6212
6314
6415
6513
6117
6222
6825
6425
6522
45
46
47
48
49
6532
6628
6721
6812
6902
6542
6637
6730
6821
6911
6551
6646
6789
6830
6920
6561
6656
6749
6889
6928
6571
6665
6768
6848
6937
6580
6675
6767
6857
6946
6590
6684
6776
6866
6955
6599
6693
6785
6875
6964
6609
6702
6794
6884
6972
6618
6712
6803
6893
6981
50
51
52
53
64
6990
7076
7160
7243
7324
6998
7081
7168
7251
7382
7007
7093
7177
7259
7340
7016
7101
7185
7267
7348
7024
7110
7198
7275
7356
7033
7118
7202
7284
7364
7042
7126
7210
7292
7372
7050
7135
7218
7300
7380
7059
7148
7226
7308
7388
7067
7152
7285
7316
7396
LOGARITHMS.
305
N
1
2
3
4
6
6
7
8
9
55
56
67
68
69
7404
7482
7559
lOU
7709
7412
7490
7566
7642
7716
7419
7497
7574
7(W9
7723
7427
7505
7582
7657
7731
7435
7513
7689
7664
7738
7448
7520
7597
7672
7745
7451
7528
7004
7079
7752
7459
7636
7612
7686
7760
7466
7643
7610
7694
7767
7474
7661
7627
7701
7774
00
61
62
63
64
7782
7853
7924
7993
8062
7789
7800
7931
8000
8069
7796
7868
7938
8007
8075
7803
7875
7945
8014
8082
7810
7882
7952
8021
8089
7818
7889
7959
8028
8096
7825
7896
7966
8035
8102
7832
7903
7973
8041
8109
7889
7910
7980
8048
8116
7846
7917
7987
8056
8122
65
66
67
68
69
8129
8195
8261
8325
8388
8136
8202
8267
8331
8895
8142
8209
8274
8338
8401
8149
8215
8280
8344
8407
8156
8222
8287
8351
8414
8162
8228
8293
8357
8420
8169
8236
8299
8363
8426
8176
8241
8306
8370
8432
8182
8248
8812
8376
8430
8189
8264
8819
8382
8446
70
71
72
78
74
8451
8513
8573
8633
8692
8457
8519
8579
8639
8098
8463
8525
8586
8645
8704
8470
8531
8691
8651
8710
8768
8825
8882
8938
8993
8476
8637
8697
8657
8716
8482
8643
8603
8663
8722
8488
8549
8609
8669
8727
8494
8555
8015
8675
8733
8600
8661
8621
8681
8739
8606
8667
8627
8686
8746
75
76
77
78
79
8761
8808
8866
8921
8976
8756
8814
8871
8927
8982
8702
8820
8876
8932
8987
8774
8831
8887
8943
8998
8779
8837
8893
8949
9004
8786
8842
8899
8954
9009
8791
8848
8904
8960
9015
8797
8864
8910
8966
9020
8802
8869
8916
8971
0025
80
81
82
83
84
9031
9086
9138
9191
9243
9036
9090
9143
9196
9248
9042
9096
9149
9201
9263
9047
9101
9164
9206
9268
9053
9100
9150
9212
9263
9058
9112
9106
9217
9269
9063
9117
9170
9222
9274
9069
9122
9176
9227
9279
9074
9128
9180
9232
9284
9079
9183
0186
0238
9289
85
86
87
88
89
9294
9346
9396
9446
9494
9299
9360
9400
9460
9499
9304
9355
9405
9456
9604
9309
9360
9410
9460
9600
9316
9365
9416
9465
9613
9320
9370
9420
9469
9618
9325
9375
9426
9474
9628
9330
9380
94:^
9479
9528
9836
0386
0435
9484
9533
9840
9390
9440
9489
9638
90
91
92
93
94
9642
9690
9688
9686
9731
9647
9696
9643
9689
9786
9652
96C0
9647
9694
9741
9667
9606
9662
9699
9746
9662
9609
9667
9703
9760
9666
9614
9661
9708
9764
9571
9019
9666
9713
9759
9576
9624
9671
9717
9763
9581
9628
0675
9722
9708
9686
9633
9680
9727
9773
95
96
97
98
99
9777
9823
9868
9912
9966
9782
9827
9872
9917
9961
9786
9832
9877
9921
9966
9791
98:^6
9881
9926
9969
9796
9841
0886
0930
9074
0800
9846
0890
9984
9078
9805
9850
9894
9083
0809
9864
0809
0943
0987
0814
9859
91K)3
9948
99*^1
9818
0863
9908
0962
0906
20
a06 ELEMENTS OF ALGEBRA.
Explanation of Table. The lefthand column, headed N, is a
column of numbers. The figures O, 1, 2, 3, 4, 5, 6, 7> 8, 9,
opposite N at the top of the table, are the righthand figures of num
bers whose lefthand figures are given in the column headed N. The
figures in the column which they head are the corresponding man
tissse of the logarithms of the numbers.
128. To Find the Logaritlmi of a Number.
I. Consisting of one Figure. The mantissae of the logarithms
of single digits, 1, 2, 3, 4, etc., are seen opposite 10, 20, 30, 40, etc.,
and in the column headed O. To the mantissa prefix the character
istic and insert the decimal point. Thus,
log 6 = 0.7782. log .6 = 1.7782. log 8 = 0.9031.
Similarly, since the mantissa of log .009 is the same as the man
tissa of log 9, log .009  3.9542.
II. Consisting of two Figures. In the column headed N look
for the figures. In the line with the figures, and in the column
headed 0, is seen the mantissa. Then proceed as before. Thus,
log 13 =1.1139. log 2.5 = 0.3979. log .92 = 7.9638.
Similarly, log .00092 = 4.9638.
III. Consisting of three Figures. In the column headed N,
look for the first two figures, and at the top of the table for the third
figure. In the line with the first two figures, and in the column
headed by the third figure, is seen the mantissa. Then proceed as
before. Thus,
• log 313 = 2.4955. log 17.9 = 1.2529. log .279 = T.4456.
Similarly, log .000718 = 4.8561.
IV. Consisting of more than three Figures, Take the man
tissa of the logarithm of the first three figures as given in the table.
Prefix a decimal point to the remaining figures of the number, and
multiply the result by the tabular * difference. Add the product to
* The Tabular difference is the difference between the two successive raan
tissjfi between which the required, or given, mantissa Ues.
LOGARITHMS. 307
the mantissa thus taken. Prefix the characteristic and insert the
decimal point as before. Thus,
1. Find the logarithm of 80.672.
The tabular mantissa of the logarithm of 806 is 9063
The tabular mantissa of the logarithm of 807 is 9069
Therefore, the tabular difference = 6
The number 80672 being between 80600 and 80700, the mantissa
of its logarithm must be between 9063 and 9069. An increase of 100
in 80600 causes an increase of 6 in the mantissa of the logarithm of
80600. Therefore, an increase of 72 in 80600 will produce an increase
of ^ of 6 (or .72 X 6), or 4.32, in the mantissa of the logarithm of
80600. Hence, the tabular mantissa of log 80672 mtist be 9063 + 4,
or 9067. Prefixing the characteristic and inserting the decimal point,
we have
log 80.672 = 1.9067.
Similarly, since the mantissa of log .0005102 is the same as the
mantissa of log 5102,
2 To find the logarithm of .0005102.
The tabular mantissa of log 510 is 7076
The tabular mantissa of log 511 is 7084
.*. the tabular diflference = 8
Hence, the tabular mantissa of log 6102 must be 7076 f .2 X 8, or
7078.
.. log .0005102 = 4.7078.
Exercise 111.
Find by means of the table the logarithms of the
following :
1. 70; 102; 201; 999; .712; 3.6; .00789; 3.21.
2. .0031; .0983; .00003; 10.08; 29461; 3015.6.
3. 32678; V337; ^/Msm2; 4^ ; (.098x85)*.
308 ELEMENTS OF ALGEBRA.
129. To Find a Number when its Logarithm is Given.
I. If the Given Mantissa is Found in the Table. The first
two figures of the required number will be seen on the same line
with the mantissa and in the column headed N, the third figure will
be seen at the head of the column in which the mantissa is found.
Finally insert the decimal point as the characteristic directs. Thus,
1. Find the number whose logarithm is 1.9232.
Look for 9232 in tbe table. It is found on the line with 83 and
in the column headed 8. Therefore, write 838 and insert the deci
mal point. Hence, the number required is .838.
II. If the Given Mantissa cannot be Found in the Table.
Find the next less mantissa, and the corresponding number ; also find
the tabular diff'erence. Annex the quotient of the difference between
the given mantissa and the next less mantissa divided by the tabular
difference, to the corresponding number ; then proceed as before.
Thus,
2. Find the number whose logarithm is 2.7439.
The next less mantissa is 7435, corresponding to 554.
The next greater mantissa is 7443, corresponding to 555.
.. the tabular difference = 8.
The diflference between the given mantissa and the next less man
tissa is 4. Since the given mantissa lies between 7435 and 7443, the
corresponding number must lie between 554 and 555. An increase of
8 in the mantissa causes an increase of 1 in 554. Therefore, an in
crease of 4 in the mantissa will produce an increase of ^, or .5, in 554.
Hence, the mantissa 7439 must correspond to the number 554+ .5, or
554.5. Therefore (II, Art. 121), write 05545 and prefix the decimal
point. Hence, the number required is .05545.
3. Find the number whose logarithm is 3.1658.
The next less mantissa is 1644, corresponding to 146.
The next greater mantissa is 1673, corresponding to 147.
.•. the tabular difference = 29.
The difference between the given mantissa and the next less man
tissa is 14. Annex \^, or .48 nearly, to the number 146, and insert
the decimal point as the characteristic directs. Hence, the number
required is 1464,8.
LOGARITHMS. 309
Exercise 112.
Find the numbers whose logarithms are :
1. 3.4683; 2.4609; 4.8055; 0.4984; 0.1959.
2. 3.6580; 2.4906; 4.5203; 2.5228; 0.6595.
3. 0.8800; 1.7038; 5.8017; 3.1144; 5.7319.
130. An Exponential Equation is one in which the expo
nent is the unknown number ; as, iif = n, ifrf = n. Such
equations usually require logarithms for their solutions.
Example 1. Solve the equation 2F = 1.5.
Process. Take the logarithm of each member, x log 21 = log 1 .6.
By means of the table, 1.3222 x = .1761.
.1761
Therefore, x = . ^^aa = .1332, nearly.
Example 2. Find the value of 3.208 X .0362 X .15734.
Process, log (3.208 X .0362 X .15734) = log 3.208 + log .0362
+ log .15734.
log 3.208 =0.5062
log .0362 =2.5587
log .16734 = 1.1969
2.2618 = log .01827.
Therefore, 3.208 X .0362 X .15734 = .01827.
Example 3. Find the fifth root of .05678.
Process, log .05678 = 2.7542.
5)2.7542 = 5)3.7542 + 5
.7508 + T = T.7508 = log .6634, nearly.
Therefore, y/.05678 = .5634, nearly.
310 ELEMENTS OF ALGEBRA.
Example 4. Find the value of log^^ 144.
Solution. To find loggi/g 144, is the same as solving (Art. 119)
(21/3)' = 144, for I, squaring each side, etc., I = 4.
Therefore, logg ^3 144 = 4.
Exercise 113.
Find by logarithms the values of the following :
1. 360 X. 0827; 117.57 X .0404 ; i^ ; (31.89)3
2. ^951; 380.57 X .000967; ^(•"^^^);^.yQ"°^^^^
212.6 X 30.2 7435 ^343
84.3 X 3.62 X .05632' 38731 X .3962' ^f2^'
4. — ■ • ^ ; 72132 X .038209 ; 7.000313.
^385.67
5. (61173)*; ^; ^; ^X^.00l; Ip.
^ ^ ' (.19268)i v'27 5^49
Solve the followiug equations :
6. 20" = 100 ; 2" = 769 ; 10" = 4.4 ; {^Y = 17.4.
7. 10^ = 2.45 ; 5''" = 2"+^ ; a/S^^^ = '^/W^.
8. 2* X 6*2 = 52* X 7'" ; 3^' = 5 ; 4" = 64.
9. (1)" =10; rrf = n\ m"*+* = 71 ; ?/i""' X c^"' = n.
10. 2^+^ = 6^ 3^ = 3x2^+^ 310^1^ = 41.^ 2^*^ = 3^^''.
11. a2*^3y = ^5 ^3x^2. :^ ^^0. ^x,^5y = (^7)4 ^^^ = (^y)8
LOGARITHMS. 311
Find the number of digits in the values of:
12. 312x28; 2^4; W^o . (4375)8. (396000)io.
Find the number of ciphers between the decimal point
and the first significant figure in the values of:
13. (.2)*; (.5y«>; (.05)5; (.0336)io ; "x/sm,
14. Given log x = 2.30103, find log xi
Find the values of :
15. loga 4 ; loga 8 ; log^ 32 ; log^ 128 ; log, 1024.
16. log2 J ; log2 J ; logj ^ ; log^ ^^ ; log^ \^16.
17. logs 729; log5l25; log, 625 ; log, 15625; log, J.
18. log_e 1296 ; log_.  ^{^ ; log. ^{^ ; logg^ 512.
19. logs ^5 125; log848 49; log8l28; loga^/s tJi
20. log,/a^^^; log27^\; logj4; log, a ; log„.
21. If 8 is tlie base, of what number is § the logarithm?
Of what J ? Of what 1 ? Of wliat 2 ? Of what 3 ? Of
what If ? Of what 2 J ? Of what 3 J ? Of what Y ^^ ?
22. In the systems whose bases are 10, 3, and J, of what
numbers is — 5 the logarithm ?
Find the bases of the systems in which :
23. log 81 = 4; log 81 =  4 ; log j^^ = 4;
log iiftW = 4; log i^ = ± 2; log 1024 = ±571.
312 ELEMENTS OF ALGEBRA.
CHAPTER XXI.
QUADRATIC EQUATIONS,
131. A Quadratic Equation is an equation in which the
square is the highest power of the unknown number.
A. Pure Quadratic Equation is an equation which contains
only the square of the unknown number; as, 5 x^ = 17.
An Affected Quadratic Equation is an equation which
contains both the square and the first power of the un
known number ; as, 5 ^^ — 2 a^ = 10.
o , ^"^ + 5 ^ a; 17
Example. Solve — I ^ = o + "t •
Process. Clearing of fractions, 12 a;^ + 60 — 9 a;^ = 4 a:^ f 51.
Transposing and uniting, x = 9.
Therefore, extracting the square root,* x = ± S. Hence,
To Solve a Pure Quadratic Equation. Find the value of the
square of the unknown number by the method for solving a simple
equation, and then extract the square root of both members.
Note. * In extracting the square root of both members of the equation
a;2 z= 9, we ought to prefix the double sign (±) to the square root of each mem
ber; but there are no new results by it, and it is sufficient to w)1te the double
sign before one member only. Thus, if we write ±, x = ± S, we have + x
= 43, ^ X = — S, — aj = j3, and — x = — 3; but the last two become
identical to the first two on changing the signs of both members. So that in
either case, x = 3, and a? rr — 3.
QUADRATIC EQUATIONS. 313
Exercise 114.
Solve the foUowiug equations :
o
3. a;(a;10) = (6a:)10; {6 x + ^f = 756^ + 5x.
35  2 a; 5x^ + 7 __ 17_Jj;
9 "*"5a,27~ 3
7. I (2 a;  5)2 = 94  24x; 3 ar^  4 = ^±j? .
8. o = o > rt ar + 7yj it' = a c^ + ??i x.
3r — n or — m
9. 2:2 ^ ^^aj_^ = ^j2:(l — wa;); 2 + 4.i'2 _. ^^(1 __^
10. 7^ — n X ■\ m = n X (x — \)', x VB + x^ = 1 + x^.
 mn — ic ?i — aa; . .^ 5
11. = ; X + vx^ — .3 =
n — mx an — x ^x^ — 3
12 ^ I ^ ^,. 1 , 1 ^^^
• 5+^ 5_a; > 1_>/1_^ l+Vla;2 a:2
314 ELEMENTS OF ALGEBRA.
132. Example 1. Solve x^ — 2ax + 4ab=: 2bx.
Solution. Transposing, we have x^ — 2ax + 4ab — 2bx= 0.
Arrange in binomial terms and factor, and we have (x — 2 a) (x — 26)
= 0. A product cannot be zero unless one of the factors is zero.
Hence, the equation is satisfied if x — 2 a = 0, or x — 2 6 = ; that
is, a x = 2ay or if X = 2 b.
Example 2. Solve ^ x2 + f x f 20^ = 42f + x.
Process. Clear of fractions, transpose, and unite,
3 x2  2 X  133 = 0.
Factor, (3 x + 19) (x  7) = 0.
Therefore, 3 x + 19 = 0, and x  7 = 0. x =  6^, and x =: 7.
Hence,
To Solve a Quadratic Equation by Factoring. Simplify the
equation, with all its terms in the first member ; then place the fac
tors of the first member separately equal to zero, and solve the simple
equations thus formed.
Exercise 115.
Solve the following equations :
1. rz;2_i0;i;=:24; a?2+2a; = 80; ^2_ iga? + 32 = 0.
2. a;2 + 10 = 13 (x + 6) ; a^ + 4x  50 = 2  5 X,
3. 4x2+13^+3 = 0; Zx^ ^ 1 = \\x  x^ ^ ^.
4. a;2ic= 11342; 5a^+3x4==8aj7rz;2_2.
5. \Zxx^ = 2x^xZ\ x^2ax\%x = \^a.
6. :i;3_5^2^5^ + 7^2. a^_^ + _9_^0.
x\Z 2xZ 3x
7. lla:2_iii =9^.
x+ 2 x1 x^2
QUADRATIC EQUATIONS. 315
8. (a;2)(a;H9a: + 20) = 0; 2x^ + Sa^^2zS = 0.
3ar— 4 3a — 2a; 4
10. mqx^ — mnx\pqx — n2y = 0; x+5 = Vx+ 5\6.
11. {ab)x^(a + b)x + 2b = 0.; x^=21 + Vx^^'
133. An aflfected quadratic equation can always be solved by the
method of completing the square. This method consists in adding to
both memlxjrs such an expression as will make the memlier, with all
the terms containing the unknown number, a perfect square. The
explanation of this methocl depends upon the principle that a trino
mial is a perfect st^uare when one of its terms is plus or minus twice
the product of the scjuare roots of the other two. This process
enables us to extract the square root of the member containing the
unknown number, and thus form two simple equations which may
be solved separately.
2x 11
Example. Solve ^(Sx)  ^^ = i (a?  2).
Process. Clear of fractions, transpose, and unite,
4x^ + 26x= 12.
Divide by 4, ar»J^a: = 3.
Add * (l^)« to lioth members, x^^^x + (J^f =  3 + (J^y = ^.
Extract the square root, x — ^z=±^.
Therefore, x^ = ^, &nd x^ = ^. x = 6, and x = f
Every affected quadratic equation may be reduced to the
general form
7Ji2^ + 7ix \ a = 0;
where m, n, and a represent any numbers whatever, positive or
negative, integral or fractional. Dividing both members by m for
a n
convenience, representing  by 6 and  by c, and transposmg, we
have
316 ELEMENTS OF ALGEBRA.
x^ \ c X = — b.
Add (2) to both members, x^ + cx+i^j =^+(2)
Or, x^+cx+ (0 =(c24i).
c
Extract the square root, x + ^ = ±^ a/c'^ — 4 6.
Therefore, x + ^ = ^ ^.c^ 4 b, and x + ^ = ^ ^c^ 4 6. From
which a: =  9 +  V^'"^ " 'I ^' ^"^ ^ = ~ 2 ~ ^ V^^ ~ ^ ^ "^^^^^
clVc246
values may be written in the form x — 5 • Hence,
Common Method of Solving Quadratics. Reduce the equa
tion to the form x^ + c x = — b. Comflete the square of
the first member by adding to each member of the equation
the square of half the coefficient of x. Extract the square
root of both member's, and solve the resulting simple
equations.
Notes : 1. * Always indicate the square of the expression to be added, in the
first member.
2. Since the squared terms of the square of a binomial are ahvays positive,
the coefficient of x"^ must be made + 1, if necessary, before completing the
square. This may be done by multiplying or dividing both members by — 1.
3. The foregoing method is called the Italian Method, having been used
by Italian mathematicians, who first introduced a knowledge of algebra into
Europe.
134. It is often convenient to complete the square without first
reducing the simplified equation to the form in which the coefficient
of x^ is 1. Thus,
3ar7 4a; 10 7
Example 1. Solve —  — + ^ _^ ^ = ^ •
Process.
1 = 7.
3X77
= +
4 X 7  1(1
7 + 5 "
h
2 + i
= i.
J =
i
QUADRATIC EQUATIONS. 317
Process. Clear of fractions,
6 a:2  1 4 a: + 30 X  70 + 8 a:2  20 ar = 7 a;2 + 35 a:.
Transpose and unite, 7 ar'^ — 39 a; = 70.
Multiply by 7X4, 196 ar^  1092 x = 1960.
Add (39)2, 196 x2  1092 x + (39)« = 1960 + 1521 = 3481.
Extiaet the si^uare root, 14 a: — 39 = ± 69.
Therefore, 14a: = 39 + 59, and 14a: = 3959. x = 7, and x =  Y
Verify by putting these numbers for x in the original equation.
3 X  V  7 4 X  V  10 _
1^ + Y + 5 ~*'
U  ¥ = I.
i = i
When a quadratic e({uation appears in the general form
»nx2 + nx + a = 0,
the terms containing x may be made a complete square, without first
dividing the equation by the coefficient of z*. Thus,
Transpose a, mx^{nx = — a
Multiply the equation by 4 m and add the square of n,
4m^x^ \ 4 7nnx \ n^ = n^ — 4a m.
Extract the square root, 2 wi x 4 n = ± ^/n^ — 4am.
Transpose n, 2 w x = — »» ± >^n^ ~ 4am
„, . — n ± \/w* — 4am
Therefore, x = . Hence,
Hindoo Method of Solving Quadratics. Reduce the equa
tion to the form in x^ + n x = — a. Multiply it hy f(/icr
times the eoefficient of x^, and complete tlu sqtiare hy adding
to each member the square of the coefficient of x in the given
equation. Extract the square root of both member s^ and
solve the resulting simple equations.
318 ELEMENTS OF ALGEBRA.
If the coefficient of x in tlie given equation is an even number, the
square may be completed as follows :
Multiply the equation by the coefficient of x^, and add to each mem
ber the square of half the coefficient ofx in the given equation.
8x 20
Example 2. Solve —7—0 "" o~ == ^*
3. "t" z o X
Process. Free from fractions,
(3 x) (8 x)20(x + 2) = 6 (3 x) (x + 2).
Simplify, 6x^5Qx = 40.
Multiply by 6, 36 x^  336 x = 240.
Add (Af )2, 36 x^  336 X + (28)'^ = 1024.
Extract the square root,* 6 a;  28 = ± 32.
Transpose, 6 a: = 28 + 32, or 28  32.
Therefore, x = 10, or  f
Verify by substituting 10 for x in the original equation.
8 X 10 20
Process. _______ ^ 6,
¥ 1 = 6,
6 = 6.
Verify by substituting — f for a: in the original equation.
8 X  f 20
Process. _^^ _____ = 6,
¥ 20 _
 4 + 10 = 6,
6 = 6.
Notes : 1. * We ought to write the double sign before the root of both
members. Thus, ± (6 x  28) = ± 32, tlie reason for not doing so is the same as
given in Art. 131, Note.
2. The Hindoo, or Indian Method, is supposed to have been discovered by
Aryabhalta, a celebrated Hindoo mathematician, and one of the first inventors
of algebra. It is not only more general in form, but much better adapted to
the solution of equations in which the coefficient of the square of the unknown
number is not 1.
3. This method has an advantage over the common method in avoiding
fractions in completing the square, and is often preferred in solving literal
equa,tions.
QUADRATIC EQUATIONS. 319
135. In case the coefficient of the square of the unknown, in
the simplified etjuation, is a square number the square may be
completed as follows :
Example 1. Solve 72 a:  54 = (20  z)(4x + 3).
Process. Simplify, 4 x^ — 5 x = 114.
""^•^ (i:Jb)' " 6)' ^ '  ^  + («' = Mj»
Extract the root, 2x i = ±^.
Transpose, 2 x = ^ + i,a. or  — ^.
Therefore, z = 6, or — 4.
The coefficient of x^ may always be made a square
number by multiplication or division. Hence,
General Method of Solving Quadratics. Add to each
member the square of the quotient obtained from dividing
the second term by twice the square root of the first term.
Tfien proceed as before.
^ , 5 3 35
Example 2. Solve —r, + 
X + 4 X X — 2
Process. Free from fractions,
5 (x  2) X + 3 (x + 4) (x  2) = 35 (x + 4) x.
Simplify,  27 x^  1 44 x = 24.
Divide by  3, 9 x« + 48 x =  8.
(48 X \*
— = ) , or (8)«, 9 x2 + 48 X + (8)2 = 56.
Extract the root, 3 x + 8 = ± 2 y/\i.
Transpose, 3x = 8±2y'14. .♦. x= r^^^
Note. The Common and Hindoo Methods of completing the square are
modifications of the General Method.
320 ELEMENTS OF ALGEBRA.
Exercise 116.
Solve the following
1. 23a?= 120 + a:2; 42 + ^2^ 13 a;; 12 ic^ + ^ ^ 1.
2. 22a:+ 232;2 = 0; a:2 rz^=i 32; 2^2 + 3 2; = 4.
3. a;+226 2;2=. 0; 25 ^=62:2+ 21; x22x = X
4. 3x'^+12l = Ux; ^ix=^x^; 91 a:2 2a; = 45.
5. 21 ^2 _. 22 ^ + 5 = ; 9 ^2  143  G a; = 0.
6. 18a;227ir 26 = 0; 50 a;2  15 a; = 27.
7. 192;= 1582;2; ^2+_4^^^1. ^2_l^_l3zzO.
8. 5 2;2 + 14 2;  55 ; (2; + 1) (2 2; + 3) = 4 2:2  22.
9. 2(^^)3(2.' + 2)(.2;3); .32^2 + 2.1 2: +  = 0.
10. 252:+22;2= 42; (2; + 6)(2;2)  f (62jV + V^)
1 __ _1 ^ ^+16 11 _ 42: 171
^^' IT^ 3^^~35' "T ■^¥~ 3
42; x — 6_4:X+7 '^Jl^ 2! — 2 _ ^
^^ "9" "^ ^+~3 ~ ~T9~"' ^~^^ '^ ^^^S ~ ^^^
.0 _J__4^__L_. _i L^1
3x 5 92 2;' 2^3 2:+5 18
14.
15.
QUADRATIC EQUATIONS. 321
4 3 4 5 3
X — 2 X X \ ^^ X — \ xf2 X
>y2+ 3 _ 12 + 5a^ 'dx 2a;5 _^,^
^^ "^aj2 5 ~ 5(0:^5)' 2:+ 1"^ 3a:l~*^*8
^^ 5a;7 a:  5 3 a:  1 ,
10. = = = ^ zTT^ ; ^ —=, = 1 —
Ixb 2a;13' 4a;+7~ x \ 1
,^ 12a:8lla;'*+ 10 a;  78 ,, ,
17. 50 — ;, — T^ = l\x — h
3a:+5 3x5 _ 135 7 21 22
3a;5 3a; + 5~176^ x^+'^x^ '6x^^x~ x
\ 18 7 8 a; ic + 3
19.
«— 1 a; + 5 a;Hl x — b' x + % 2a;+l
136. Literal Quadratic Equations.
Example 1. Solve mx"^ \nx = — ; — ^ m x  n x^.
Procesa. Transpose and factor, (m 4 n)x^— {m — n)x = — — •
Multiply the equation by 4(m4 n) and add the square of (m — n)^
4 (m + nyx^  4 (m2  n^) a: + (m  n)« = (m + n)«.
Extract the square root, 2(m}n)a:— (m — n) = ± (to + n).
Transpose, 2 (to + n) a: = to — n ± (m + »»)
= 2 TO, or — 2 n.
«« * TO n
Therefore, x  , , or
TO + n' TO + n
« « , 2a:+l 1/1 2\ 3x4 1
Example 2. Solve — r ( r — ~ ) = — :: —
x\o aj a
21
322 ELEMENTS OF ALGEBRA.
Process. Free from fractions,
ax(2x+l) (a2b) = bx(Sx + l).
Simplify and transpose,
2ax^i ax 3bx^~bx = a~2b.
Express the first member in two terms,
(2 a  3b) x^ + (a  b) X = a  2b.
Multiply by 4 (2 a  3 b),
4 (2a36)2a;2+4 (2a36) (ab)x = 8 a^28ab + 24 b\
Complete the square,
4(2a36)2a;2+4(2a36)(a6)2: + (a&)2 = 9a230a6 + 25 62.
Extract the square root,
2{2a  Zb) X + {a  b) ~ ± {3 a  bb).
Transpose, 2 (2 a  36) x =  (a  6) ± (3 a  5 6)
= 2a4&, or2(2a36.
a2b
Therefore, x = ^^^g^ , or  L
1111
Examples. Solve ^ + ^"i;^ =+ ^ ^ ^
111 1
Process. Transpose,   " = ^+6 " j+^
Reduce each member to a common denominator,
a — X X — a
ax ia + b)(b + x)
Free from fractions, {ax){a\b){b + x)= ax(x a).
Transpose and factor,
(a  X) [(a + b)(b + x) + ax'] = 0.
Hence (Art. 11), a  a; = 0. .',x = a.
Also, (a\b)(b + x) + ax = 0.
Simplify and factor,
^ ^ b(a + b)
b(a + b) + (2a + b)x = 0. .'. x=  ^^^^ ■
Kote. Always express the first member of the simplified equation in two
terms, the first term involving x'^, the second involving x.
QUADRATIC EQUATIONS. 323
Exercise 117.
Solve the following equations :
. 9 / . r\ . 7 f\ 2 . ^^^ m ax
1. x^ — (a{b)x + ab = 0: mx^] m = — 7 — .
^ X a X b ^ mbx am z
2.  +  =  + ; ma^ m = r
a X b X a
3. (a6)r^(a + 6)a;+26 = 0; a^ 3^ ^ abx=^ 21?.
7^ X ^2a^ 2 z(a — x) _ a a^ + m2_
a^'^ b^~W' %a2x ""V ^ ~
f>, ^ — n X •\ 'p X ^ n'p ■=■ ^ \ a^ \ 2x Vn = n.
6. a^3^ ~2a^x+ n* 1 = 0; a^ \ x (a  b) = ab.
7. x^\mx^cx\Ji^x+m^x = 0.
8. ^a^x = (a^b^ + xf; a^ {x  a)^ = 1)* (x + a)^
\a X J \ X a J
10 —i . rxQ = (« — ^)^ ; ^— ^ X — X = —CX — Z.
{a\r by ^ ' b n
11. («'~^(^J+l) ^2a?; 9a*6*2^»~6a8 63^ = 62.
19 ^^+ <^ _ ^46 11 1 _^
« — 6~~na; — a' a a + a; a + 2a7~~
324 ELEMENTS OF ALGEBRA.
ax X b (al)^s^ { 2(Sa^l )x _^
lo. r ■ " — —  ; ;; z — 1.
X a — X c 4a — 1
14. ra; + ^y = 4a:2; (ax'^=\a?x\
137. Solution by a Formula. From the quadratic equa
tion moi?' {■ nx — ^ a,
— n± Viv^ — 4:am , ^ .
*= 2Vv (^>
By means of this formula the values of x, in an equation
of the general form, may be written at once. Thus,
Example L Solve 10ic2_ 23 a: =  12.
Process. Here, m = 10, n = — 23, and a = 12.
c V .. . .u 1 • /ix (23)± V(2 3)^ 4X 12X 10
Substitute these values m (1), x = ^ y 10
23 ±7
~ 20
= i or f
^ r. , 1 111
Example 2. Solve r ; — = j +  h
b + c \ y c y
Process. Free from fractions, transpose, and factor,
(h + c)y^+ (h + cyy = hc{h[ c).
Divide by 6 + c, y^ ^ (h+c)y = — he.
Here, m = 1 , n = 6 + c, and a = hc.
Substitute these values m (1), x = ~
(b + c)±(bc)
~ 2
= — r, or — 6.
Note. In substituting the student must pay particular attention to the
signs of the coefficients.
QUADRATIC EQUATIONS. 325
Miscellaneous Exercise 118.
1. 17 a^ f 19 a* = 1848; Sa^  12x + 1 = 6x2S.
2. 5a:2 + 4^. = 273; ^x' + lx + ^ = 0,
^•^+2^=& «(^' + 3)«=(.+ 3)^f
4 x + 1 _ a;+2 _ 2a; + 16 x2
'^'^'^bx'^ 5 ~^' xl~ x + 5 x + 1
2a;+3 7 — x 7—3x
5. 16x^6x'l = 0;
2(2a:l) 2(a:+l) 43a;
a 5x_^ xS 2 (a: + 8) ^ 3 a: + 10
3"^'4~3a^a:5"'" a:44 ~ a:+l
7.^=^; H(5^ + 36)2 = V^j(8a:24)^.
X CL
8. 4 =  + ; lla?»410aa: = ±a2; a:' +  + 2 = 
p\x p x' 'XX
, ,2:2j2.fl,2fltJ
9. ax ■\ dx^ — a= d\ — « H = — o H
ir^ c 7?r c
10. WX2
(w^ — w^ a; _ a^ a _
mn '3m — 2a 2 4a — (3 w
11. a:^  2 a X = (6  c + a) (6 — c — a).
12. x^  (a { b)x = \{7n \ n \ a \ h){m ^ nah).
,o 1 _^ 1 a2 + a:2 ^^3 ^r  3
a k X a — X a*  or x — 6 x \ .*
326 ELEMENTS OF ALGEBRA.
14. ahx'^  2x{a + h) ^/'^ = {a hf.
' ^' "^ aW' ~ 18aH2 + 2ah
a — 2h~x 5h — x 2 a — x — 19h_
a^ — 4:b^ ax + 2bx 2bx — a x ~~ '
^^1 1 m 2nx+ n
^7 I . =:
2x^{x~l 2^2— 3 a; +1 2nx — n mx^ — m
18. = ax.
ax — Va^ x^ — \ ax + \o? x^ — 1
Query. What is the diflference between the meaning of "the
root of an equation" and "the root of a number"?
138. Problems. The following problems lead to pure or affected
quadratic equations of one unknown number. In solving such prol)
lems, the equations of conditions will have two solutions. Some
times both will fulfill the conditions of the problem ; but generally
one only will be a solution.
Exercise 119.
1. Find a number whose square diminished by 119 is
equal to 10 times the excess of the number over 8.
Solution. Let x = the number.
Then, x — 8 = the excess of the number over 8.
Therefore, a:^  119 = 10 (a:  8).
The solution of which gives, a: = 13, or — 3.
Only the positive value of x is admissible. Hence, the number
is 13.
Note. In solving problems involving quadratics, the student •should retain
only those values for results that will satisfy the conditions of the problem.
PROBLEMS. 327
2. The difference of the squares of two consecutive
numbers is 17. Find the numbers.
3. Find two numbers whose sum is 9 times their differ
ence, and the difference of whose squares is 81.
4. Find two numbers, such that their product is 126,
and the quotient of the greater divided by the less is 3 J.
5. Divide 14 into two parts, such that tlie sum of the
quotients of the greater divided by the less and of the less
by the greater may be 2r^.
6. Find two numbers whose product is m, and the
quotient of the greater divided by the less is n.
7. Find a number which when increased by n is equal
to m times the reciprocal of the number. Find the num
ber when n = 17 and vi = 60.
8. Divide m into two parts, so that the sum of the two
fractions formed by dividing each part by the other may
be 71. Solve when m = 35 and n = 2^,
9. Divide a into two parts, so that n times the greater
divided by the less shall equal in times the less divided by
the greater. Solve when a = 14, ?i = 9, and m = 16.
10. A farmer bought some sheep for $72, and found
that if he had received 6 more for the same money, he
would have paid S 1 less for each. How many did he
buy?
11. If a train travelled 5 miles an hour faster, it would
take one hour less to travel 210 miles. Find the rate
travelled and number of hours required.
328 ELEMENTS OF ALGEBRA.
12. A man travels 108 miles, and finds that he could
have made the journey in 4 hours less had he travelled
2 miles an hour /aster. Find the rate he travelled.
13. A number is composed of two digits, the first of
which exceeds the second by unity, and the number itself
falls short of the sum of the squares of its digits by 26.
Find the number.
14. A number consists of two digits, whose sum is 8 ;
another number is obtained by reversing the digits. If
the product of the two is 1855, find the numbers.
15. A vessel can be filled by two pipes, running to
gether, in 22 minutes ; the larger pipe can fill the vessel
in 24 minutes less than the smaller one. Find the time
taken by each.
Solution. Let x = the number of minutes it takes the larger pipe.
Then, x \ M = the number of minutes it takes the smaller pipe.
 =r the part filled by the larger pipe in one minute,
and ^^ = the part fi lied by the smaller pipe in one minute.
r^, P 1 1 1
Therefore,  +
X ' a; f 24 ~ 22f
The solution of which gives, x = 36, or — 15.
One pipe will fill it in 36 minutes, and the other in 1 hour.
16. A vessel can be filled by two pipes, running to
gether, in m minutes ; the larger pipe can fill the vessel in
n minutes less than the smaller one. Find the time taken
by each. Solve when w = 56 and w = 66.
PROBLEMS. 329
17. B can do some work in 4 hours less time than A
can do it, and together they can do it in 3 hours. How
many hours will it take each alone to do it ?
18. A boat's crew row 7 miles down a river and back
in 1 hour and 45 minutes. If the current of the river is
3 miles per hour, find the rate of rowing in still water.
19. A boat's crew row a miles down a river and back.
They can row m miles an liour in still water. It took n
hours longer to row against the current than the time to
row with it. Find the rate of the current. Solve when
a = 5, w = 6, and n = 2.
20. A uniform iron bar weighs m pounds. If it was a
feet longer each foot would weigh n pounds less. Find
the length and weight per foot. Solve when m = 36,
a = 1, and n = .
21. A and B agree to do some work in a certain num
ber of days. A lost m days of the time and received n
dollars. B lost a days and received c dollars. Had A lost
a days and B m days, the amounts received would have
been equal. Find the number of days agreed on and the
daily wages of each. Solve when m = 4, 71 = 18.75, a = 7,
and c = 12.
22. A pei"son sold goods for vi dollars, and gained as
much per cent as the goods cost him. Find the cost of
the goods. Solve when m = 144.
23. By selling goods for m dollars, I lose as much per
cent as the goods cost me. Find the cost of the goods.
Solve when m = 24.
330 ELEMENTS OF ALGEBRA.
CHAPTEE XXII.
EQUATIONS WHICH MAY BE SOLVED AS QUADRATICS.
139. In the equation m {if — yy + n (y^ — z/)^ f a = 0, suppose
(y^ — VY = 'C, then mx^ + nx la = 0. Similarly, y^Sy^ — 9 =
may be changed to the form a:^ — 3 a: — 9 = 0.
Hence, an equation is in the quadratic form when the unknown
number is found in two terms affected with two exponents, one of
which is twice the other ; as, a;^ + 5 a;^ — 8 = 0.
The general form for an equation in the quadratic form is,
ax^'' + b^f' + c = 0;
where a, h, c, and n represent any numbers whatever, positive or
negative, integral or fractional.
Example I. Solve a;*  13 x^ + 36 = 0.
Process. Factor, (x + 2) (x  2) (x + 3) (ar  3) = 0.
Hence, a; + 2 = 0, a;2 = 0, a; + 3 = 0, and x — 3 = 0.
Therefore, a; = ± 2, or ± 3.
Example 2. Solve 8 a;~ ^  15 x~^ 2 = 0.
Process. Factor, (x~^ — 2) (8 a:~^ + 1) = 0.
^i_2 = 0, ora:^ = i x = (hS^ = i ^2.
Also, 8 X ^ + 1 = 0, or a;' =  8. x= ( 8)^ =  32.
Example 3. Solve 3 a: + a:^  2 = 0.
Process. Solve for x^. Thus,
Multiply by 12 and transpose, 36 ar + 12 a:* = 24.
Complete the square, 36 a; 4 12 a:^ + 1 = 25.
Extract the square root, 6 a:* + 1 = ± 5.
Therefore, a;* = f , or  L
Square each member, a: = ^, or 1.
EQUATIONS SOLVED AS QUADRATICS. 331
Example 4. Solve 2 ^^/^^ ^ 3 ^^ _ 55 _ q
Process. Since ij/x^ is the same as x~ ', and /y/x*^ is the same
as x~ *, this equation is in the quadratic form. Transpose and mul
tiply by 12, 36 X* + 24 x~^ = 672.
Complete the sciuare, 36 z"* + 24 x~* + 4 = 676.
Extract the square root, 6 a:~ * f 2 = ± 26.
x~^ = 4, or — ^.
Therefore, a:' = J, or — ^.
Extract the square root, a:* = db ^, or ± \/~ ^,
Raise to the 5th power,* x = ± jij, or ± V(~A)*
Hotes : 1. When the roots cannot all be obtained by completing the square,
the method by factoring should be used. Thus, in solving a;* + 7 a:* — 8 =: 0,
by completing the square, we find but two values for x, re = 1, or — 2. Fac
toring the first member, we have (x + 2) (a* — 2 x + 4) (x  1 ) {x^ + x\l) = 0.
Hence, a; + 2 = 0, a^ _ 2 a: + 4 = 0, a;  1 = 0, and x^ f a: + 1 = 0. Solv
ing
these equations, a: = — 2, 1 ± V^^, 1, and
2. • In solving equations of the form a;» = a, first extract the wth root,
and then raise to the »ith power. In practice this is the same as affecting the
•quation by the exponent — . Thus, a: = a« .
m
Example 5. Solve a a:^" + 2» jr" = — c.
Process. Multiply by 4 a,
4 a* x^'* + 4a6ar"=: — 4ac.
Complete the square,
4a2j:2" + 4abx'* + b^ = 4ac+b^.
Extract the square root, 2 a a* + 6 = ± yb^ — 4ac.
Transpose b and divide by 2 a, af = — ^ — •
Extract the th root, x = [±VE^ILz*]" (i)
Example 6. Solve Ax*  37 x' + 9 = 0.
332 ELEMENTS OF ALGEBRA.
Process. Here, a = 4, 6 = — 37, c = 9, and n = 2.
Substitute these values in(i), x = [ ^ V(37)^4X4X 9(37)f
^^' L 2X4 J
r ± 35 + 37 1^
= ± 3, or ± i
Exercise 120.
Solve the following equations :
1. 0^414^2^40; a;io + 312^5 = 32; x^7x^ = S.
2. 2^(19 + :i^3):^ 216; 2^2 + ^ ^.^ ,,2 + ^2
3. 16(':c2+ i^ = 257; ^3 +14^^1107.
4. 5 a:* + \/x = 22 ; 'v/^ +  = SJ.
2 V 2
5. ^t + 7 ^t = 44; 3 a;3 + 42 x^ = 3321.
6. x^ + x^= 756 ; 3 V^^ _ 4 ^.^ = 7,
^ . .. _2 . 2
Vx ' xi
7. 2V^ + ^^ = 5; 122:t + f = 4 + ¥
a 3 ^t  a; + 2 = ; 2 a;5 + 61 a^t  96 = 0.
9. x^ + axi = 2a^; x^2x^ = S', x^ + V^=(^
10. rr*"  2;2»   = ; 3 ict'* + 4 xl"" = 4.
11. a:" + 13 o:''* = 14 ; 3 :r '"^  26 x ^'" = 16
EQUATIONS SOLVED AS QUADRATICS. 333
140. ?>iuatioijs may frequently be put in the quadratic form by
grouping the terms containing the unknown number, so that the
exponent of one group shall be twice the exponent of the other group,
and then solved for the polynomial. Thus,
Example 1. Solve a;  3 x*  4 Va:  3 x*  1 =  2.
The aquation may be put in the quadratic form if we reganl
Vx — 3 a:* — 1 as the unknown number. Thus,
Process. Add — 1 to each member,
x3a;*l+4Vx3ar*l=3.
Put Vx 3x^1 = y, y^ + 4y = 3.
Therefore, y = 3, or 1.
Hence, Vx  3 x*  1 = 3, or 1.
Squaring, ' x  3 x* — 1 = 9, or 1.
Complete the square, x  3 x* + } = ^, or Jf .
Solving these equations for the values of x, we find x = 25, or 4,
13±3v^
and X = ^ — —
Hote 1. In solving equations of this fonn we must group the terms so that
the expression outside of the radical, in the first member, is the same or a mul
tiple of the expression under the radical sign.
Example 2. Solve x*  6 ax« + 7 a" x^ 4 6 a»x = 24 a*.
Process. Add 2a«x^ x*6ax^+9a^x^^6a*x = 24 a* f 2a«x«.
Transpose 2 a«x2, x* 6 a x«+ 9 a^ x2+ 6 a«x  2 a^x^ = 24 a*.
Group and factor the terms,
(x2  3 ax)2  2 a2 (x2  3 ax) = 24 a*.
Regard x*— 3 ax as the unknown number, and complete the square,
(x2  3 a x)a  2 a« (x2  3 a x) f a* = 25 a*.
Extract the square root, (x'^ — 3 a x) — a^ = ± 5 a*
Therefore, x«  3 a x = 6 a", or ~ 4 a«.
Complete the square and solve, x = 2 (3 ± \/33),
334 ELEMENTS OF ALGEBRA.
Note 2. Form a perfect square with xi and —6a x^. The third terra of the
square is the square of the quotient obtained by dividing 6ax^ by twice the
square root of x*.
Example 3. Solve a:2 + 4x — 4a:rifx2 = ^.
Process. Use positive exponents, rearrange terms, and factor,
a=^ + ^, + 4(xi)=.
Regard x as the unknown number, and subtract 2 from both
sides, „ ^ 1 . f 1\ 11
Factor, and complete the square,
(.iy+4(.^)+4=^.
Extract the square root, a: —  + 2 = ±f.
Therefore, x —  = — ^, ot — ^.
X
Free from fractions, ^ ar^ — 1 = — ^ a:, or — ^ x.
Complete the square and solve, x = ^ (— 1 ± y'S?) ,
x = ^(ll±^/T57).
Note 3. Form a perfect square with x^ for the first term and — for the
third. The middle term will be twice the product of their square roots taken
with a negative sign.
A Biquadratic Equation is an equation of the fourth de
gree. Biquadratic means twice squared, and hence the
fourth power.
If a biquadratic is in the form,
x^+2mx^+ (m2 + 2 n) a:^ + 2 mnx = a (ii)
the first member becomes a perfect square by
Adding n^, or the square of the quotient obtained by dividing the
coefficient of x by the coefficient of x^.
EQUATIONS SOLVED AS QUADRATICS. 335
Thus, extracting the stiuare root of the fiist member,
X* i 2mx^ + {inr +'2,n)x^ \ 2mnx  x'^ + mx { n
X*
2x^\ mx I 2mx^+ (m^ + 2n)x^\2mnx
2mx*+{m^ )x^
2x^ + 2mx\n\ + ( 2n)x^ + 2mnx
+ ( 2n)x^+ 2mnx + n^
— n*. Hence,
the equation may be written,
{x^ \mx + n)2  n^ = a, or (z" + w z + «)« = a + n^ (iii)
Example 4. Solve x*  10 «« + 35 z^ _ 50 x = 1 1.
Process. Here, 2m = — 10, 2 win = —50. .. m = — 5 and n = 5.
Since m^ + 2n = 35, the equation has the form of (ii).
Add 25 ; or put w = — 5, n = 5, and a = 1 1 in (iii),
(z2  5 z + 5)2 =r 36.
Extract the square ^oot, z* — 5z + 5 = ±6.
Therefore, z^  5 z = 1, or  11.
5 ± a/29
Complete the square and solve, z = ^ — ,
5± a/^Hq
^ = 2—
Kote 4. After adding the value for n* the first member may be factored by
substituting the values for m and n in (iii).
Exercise 121.
Solve the following equations :
1. (3^ + x'2)^lS(2^ + x2) + Z6 = 0.
2. a^» 4 2 a; + 6 Va^+2x+5 = 11.
3. 2:2 + 24 = 12 V^qrf. 2a: + 17 = 9 V2 a;  1.
4. a:2  a: + 5 (2a:2  5 a; + 6)i = J (3 a: + 33),
336 ELEMENTS OF ALGEBRA.
^ (a;2 _^ ^,. ^ 6)i ^ 20  (a;2 + a; + 6)^
7. (« + l)%4(«+!) = I2.
8. (a;2  5 a;)2  8 (ic2 _ 5 ^) ^ 43^
9. 9 ^  3 ^2 _!_ 4 (^2 _ 3 ^ + 5)^ ^ il^
■«(9"!(^)S
11. (3a:210^+ 5)28(3^ 10a^+ 5) = 9.
13. :i:4 + 6^^ + 5:z;2 12a;=12.
14. x^Qa^2^x^+ 114 2; = 80.
15. ^4 + 2 2^  25 a;2 _ 26 a; + 120 = 0.
16. 2:48 .>;3 + 10 2:2 + 24 a: + 5 = 0.
17. a:4 + 8 2:3 + 2 a:2  I a: = _
18. (^3 _ 16)1 _ 3 (2^3  16)i = 4.
19. ^Y^^Vxx^^A.; 2;2+3:^32:i + 2r2 = ^.
EQUATIONS SOLVED AS QUADRATICS. 337
141. Equations Containing Radicals may be Solved. Thus,
Example 1. Solve x  ^3* + 2 x + 12 + 2 = 0.
Process. Transpose, x + 2 ^x» + 2x f 12.
Raise both members to the third power,
a:» + 6 x» f 12 X + 8 = a:* + 2 X h 12.
Transpose and simplify, 3 a:"^ + 5 x — 2 = 0.
Factor and solve, x = ^, or — 2.
Verify by putting these numbers for x in the original equation.
Process, x = J. x =
2  ^84+12 + 2 = 0,
20 + 2 = 0,
= 0,
1 1
i  ^tjV + f + 1^ + 2 = 0,
i  i + 2 = 0,
= 0.
Example 2. Solve
Vx* + 1 ^/x'^  1 ^/J^  1
Process. Multiply by /^/x* — 1,
Vx2  I + yx« +1 = 1.
Square, x^  1 + 2 ^/x'^~^\ + x« + 1 = 1.
Transpose and simplify, 2 ^/x* — 1 = 1 — 2 x^.
Square, 4a:*4 = l4x« + 4x*.
Simplify, x« = f .
Extract the square root, x = i J y'S.
Exercise 122.
Solve the following equations :
1. 3 V^+6 + 2 = a;+VT+6; a:+ \/iT~2 = 10.
2. aj 4 16  7 v./ + 16 =104 Vx + 16.
3. 2a:+ V4a; + 8 = J; V4a;+ 17+ V^l » 4.
338 ELEMENTS OF ALGEBRA.
4. 2'v/3a; + 7 = 9V2i^3
 V4:X + 2 4  Vi
4A/ic Vx
12 5x 9 _ \/5^3
Vic+ 12' V5^+3 2
442;
5. V^ + ^ = y ; 7= — ;==! +
Vic + 12 V a; + 3
6. Vi^2\/^=2;; 1^64 {2x^Sx ,,
V4: \ X
H, » A/ . * A/ o /o— 3.a; — 1 . , Vi^a; — 1
+ ^ = a;.
X + V'2 — x^ X — V2 — x^
^ V7 y2 + 4 + 2 a/3 7/  1 ^ m Vrn^  y^
a/7 2/^ + 4  2 A/3y 1 ' ^^^ + A/m2  2/2
10. ^a2 + 2a^2_2aaj = ^^iL; A/6a:a;3= "^
11.
Va + .^' A/a;
a^&2 V^ + & a/.^: + 9 3a/^3.8
a/^ 2; + & ^ ' Vx 9 — Vic
^^ /—  / 12a 6 + SVx 4
12. Va + X {■ ya — x= ; 7— r — 7=^ = "7=
5 A/a + a; 4 + Y^,^ Va;
13. V^f^+Jj  Vy^^ = V2y; 2rr+3A/^=27.
, , a; + a/S 2:22; 12 + 8 a/S
14 ^ — — — • X — z
X Vx 4 ' xb
o_^ + ^'^^. a; _ A/a; — 12
2; '4 2: — 18
16. A/2;^ + A/2r^ = 6 a/5; Vx — a ■\ "sJx ^ a = ^/'^
THEORY OF QUADRATIC EQUATIONS. 339
THEORY OF QUADRATIC EQUATIONS.
142. Representing the roots of mx^ + nx = ~ a by r and r^, we
have (Art. 134),
— n + \/n^ — 4 a m ^
— n  \^n^ — 4am
'1 'Zm
n
' + '. = »
(i)
a
rr, — —
*■ m
(ii)
Adding,
Multiplying,
Hence, if a quadrntic appears in the form, mx^ + nx = — a^
I. The sum of the roots is equal to the quotient ^ with its sign changed,
obtained by dividing the coefficient ofx by the coefficient ofx^.
II. The product of the roots is equal to the second member, with
its sign changed, divided by the coefficient of x^.
By means of (i) and (ii) the ori«,Mnal equation becomes,
m x* — wi (r f r,) a: 4 m r Tj = (1)
Factor, m(xr)(x  r,) = (2)
If m = 1, x^\nx = — a
x^ (r + rj) X + rrj = (iii)
ixr)(xr,) = (3)
If the roots of a quadratic equation be given, by means
of (iii) we can readily form the equation.
Example 1. Form the equation whose roots are ^, —\.
Process. Here, r = ^ and r^ = ~ \.
Substitute these values in (iii), x^ (^\)x + (\) (\) = 0.
Simplify, 8 x«  2 x  1 = 0.
Example 2. Find the sum and the product of the roots of
8x«+3x5 = 0.
Process. Here, m = 8, n = 3, and a = — 6.
Substitute in (i) and (ii), r + r j = — { and rrj = — j.
340 ELEMENTS OF ALGEBRA.
Exercise 123.
Find the sum and product of the roots of :
1. 2:2 + 8^ = 9; 12 a:'^  187 ^" + 588 = 0.
2. 20x^ = 55x^l a^6x+9 = 9x.
3. 32:2 + 5 = 0; ^24.^2:=^^. a^l5x = S.
. „ 2mn^x mn 207, 2 . 1.2 n
4. x^ = ; x^ — 2 X — 0? \ h^^ = 0.
m — n m — n
Form the equations whose roots are :
5. 7,3; ,; 5, 3; ± V=^; 2 V3, 2 + V3.
6. 0,5; 7 + 2A/5, 72a/5; 1 + V2, 1  V2.
7. 7/1 (m + 1), 1 — //I ; — , ; 1 , — a.
^ ^ n m a — h
8.  w + 2 \/2 /I,  w  2 V2 71, > ^—
143. A Root is said to be a Surd when it can be found
only approximately ; as, a; = =t ^^,
Real Roots are values of the unknown numbers that can
be found either exactly or approximately.
Imaginary Roots are values of the unknown numbers
that cannot be found exactly or approximately; as,
x = ± V^^.
Character of Roots. For brevity, represent the roots of the
equation mx^^nx + a by r and r^, then,
r= TJ}L ,
2m
_ — n — \/7i^ —4 am
^^ 2m "'
THEORY OF QUADRATIC EQUATIONS. 341
It is seen that the two roots have the same expression, y/n^—Aam.
If n^ is greater than 4 am, n* — 4 a m will be positive^ and
\/n* 4am can be found exactly or approximately.
If n is positive, r^ is numerically greater than r ; if n is negative,
r is numerically greater than Tj. Hence,
I. Condition for Eeal and Different Ebots. n*  4 a m,
positive.
nitiatration. 3a:22x +  = 0.
Here, wi = 3, n = — 2, and a = .
n24am= (2)2 4X^X3=41 = {.
Therefore, the roots are real and different.
Evidently both roots will be rational or both surds according as
n* — 4 a m is, or is not, a perfect square. Hence,
11/ Condition for a Rational or a Surd Root, n^ 4am,
a square number; or, /y/n — 4a7/t, a surd.
lUustrationB. (1) z*  3 x  4 = ; (2) 8 x^ + 5 x  J = 0.
(1) Here, m = I, n =  3, and a = — 4.
n2  4 am = ( 3)*  4 X  4 X 1 = 9 H 16 = 25.
Therefore, the roots are real and rational, and dilferent.
(2) Here, m = 8, « = 5, and a — — \.
Vw* 4am = \/25 + 8 = \/33
Therefore, the roots are real and surds, and different.
If n* is less than 4am, n'— 4am will be negative, and \/«*— 4am
will represent the even root of a negative number. Hence,
III. Condition for Imaginary Roots. n«4am, negative,
Uluatration. 2x«3x + 2 = 0.
Here, m — 2, n = — 3, and a = 2.
n«4am=(3)« 4X2X2^9 16 = 7.
Therefore, the root.<< are both iniM^'inary.
If n* = 4am, n*4«m = 0, and the roots will be real and equals
and have the same sign, but opposite to that of n. Hence,
342 ELEMENTS OF ALGEBRA.
IV. Condition for Equal Roots, n^ ~4am = 0.
Illustration. 4x2 — 12a; + 9 = 0.
Here, m = 4, n = — 12, and a = 9.
71^ 4 am =144 144 = 0.
Therefore, the roots are real and equal.
If a m is positive, for real roots, n^ — 4am will be positive and
less than n^, since ^n^ — 4am will be less than n.
If a 771 is negative, ^n^ — 4 am will be greater than w, since
n^ — 4 a m will be greater than n^. Hence,
V. Condition for Signs. // a m is positive, real roots have the
same sign but opposite to that of n. If am. is negative, the roots
have opposite signs.
Illustrations. (1) 2x210a;+12 = 0; (2) 2x^5x3\^ = 0.
(1) Here, m = 2, n =  10, and a= 12.
n2  4 a m = 100  96 = 4.
Therefore, the roots are rational and positive, and different.
(2) Here, m = 3, n = — 5, and a = — S^J.
n2 4 am = 25 + 47 = 72.
The roots are surds and have opposite signs, and different.
Exercise 124,
Determine by inspection the character of the roots of :
1. 5x'^x = 3; 7x^+2x=^: Ux^x^^,
2. 4^2+ 52a; = 87; 3 x^ + 4a' + 4 = 0.
3. 6llx9x'^ = 0; 9a = 3 + 4la;2
4. 10x+S:^ = 3x^ lx'^^x + l=0.
5. 3a:22a:+3=:0; 4:X^3x5 = 0.
THEORY OF QUADRATIC EQUATIONS. 343
8. 6ar2+ 5 2:21 = 0; 13 ar^ + 56 a;  605 == 0.
9. 9 ar^  30 a; + 41 = ; 40 o,^  100 a;  360 = 0.
Query. How many roots can a quadratic equation have ? Why ?
Miscellaneous Exercise 125.
Solve the following equations :
1. a:t + 7J = 44; x^2x^ = S; 3s^:^2 = 0,
2. k/t^ + J = 2X 1 + 8 a:* h 9 '>y^ = 0.
▼ i — X ^ X
3. —== — 2 V2 a: = 59 ; ., ^ , ._ + — = = 3 Va;.
^/2x 1 + 5 Va; Vx
4. a;*"2a:3« + ^ = 6. 3^2o(^ + x = a.
5. a^ + ^a:3_39^ = 81. a^ _ 2 a^ + a; = 380.
6. 108a:*= 180^8 20aj Sla^H 7.
7. a:*10a:2 + 35 a:^ _ 50^; =  24.
8. {x  rt)f + 2 v^ (a; ~ rt)i  3 7i = 0.
9. a:t ~ 4 a:f + r J + 4 a:f =  {.
344 ELEMENTS OF ALGEBRA.
V y \ 2 a — ^y — 2 a _ y x \ ^x _ o^ — x
^y2a + V2/ + 2a ~" 2 a' x  '^x 4
4:r"
11. 3a;"'^a;"^:=:4; V'6:r+l + K2: + 4+ V^^+l = 2.
12. a; V5 + a/2 a' 4 2 = V^ + :^; a:  1 = 2 + —  •
13. 2 Vi + 2 :z^i = 5 ; 6 Vi = 5 :c 2  13.
14. x^ + 2 m^x^ = ^m; oT^ + 2 = ^ ,"^ •
X 3 + 5
15. ^^2;+ V2^^=TKx V2^T = I v/ — ^^^=.
ic
+ Va^  1 a:  V:r2  1
16. ^ ^ ^ ^  :: ^ = 8 ^' Va:^  3 ^^ + 2.
a; — Vx^ — 1 2: + Va;2 — 1
17. — ^^ — ^~^, ^^— = & ; ic3 + a; A/a;  72 = 0.
ft + 2;
18. State the conditions that will make the roots of
x^ + Ax + B = 0: (i) surds; (ii) real; (iii) imaginary;
(iv) equal; (v) have same signs; (vi) have opposite signs;
(vii) equal in value but opposite in sign.
19. Find a number such that if its nth root be increased
by one half of its ^th root, the sum shall be a. Solve
when n = 2 and a = 5.
20. Find a number sucli that if its nth power be dimin
2 a . '
ished by the  th root of the th part of it, the remainder
' n c ^
shall be m. Solve when m = 144, n = 2, a = 27, and c = 5.
SIMULTANEOUS QUADRATIC EQUATIONS. 345
CHAPTER XXIII.
SIMULTANEOUS QUADRATIC EQUATIONS.
144. Only certain forms of quadratic equations involving two
unknown numbers can be solved. Thus,
Example 1. Solve the equations : ^ ^ ^„ ^~ « « ^. ^2
10  1/
ProcesB. From (1), x= ^ (3)
/l()_y\2 /10w\
Substitute in (2), 2 ( — ^^ j  ( "^ ) 2/ + 3 3^« = 54.
Simplify and factor, (y — 4) (4y + 1) = 0.
Therefore, y = 4, or — ^.
Substitute in (3), ar = 3, or 5^. Hence,
When one of the Equations is of the First Degree. Solve
by substitutiou.
The Degree of a term is the mimber of literal factors
involved, and is always equal to the sum of their ex
ponents.
Each literal factor is called a Dimension.
Thus, 3 xy is of the second degree, and has two dimensions.
5 x«j/* is of the fjih degree, and has Jive dimensions.
« o 1 .1 *• 5l83xy + 72x+36y = 88 (1)
Examples. Solve theequations :  ^^^^^^^^^^3^^ ^ g^ ^^^
8036y .^x
Process. From (2), a;  Y77— jTgo ^' ^
Substitute in (1),
183(80y36y«) 72(8<l36y) , _„ _ ^^
177y + 60 + 177 y 460 + ^^ ^ " ^^^
Simplify, 9 y* + 57 y  20 = 0.
Complete the square and solve, y = J» or — 6f .
Substitute in (3), ar = ^, or  f
346 ELEMENTS OF ALGEBRA.
Example 3. Solve the equations .  6 x^  x  3 3/ = 5 (1)
^ lx^ + xy=l (2)
Process. From (1), y = :: (3)
Substitute in (2) and simplify,
3a:24a:2 = 0. _
Complete the square and solve, x = :z
^ , . . , . 11 ±7 ViO ,,
Substitute m (3), y — ^ . Hence,
When each Equation Contains only one Second Degree
Term, and that Term Consists of the Same Product or Square
of the Unknown Numbers. Solve by substitution.
Exercise 126.
Solve the following equations :
C^xy = 50. C2x + y = 22.
2. 3.;j7/+6^2z/ = 4. g {xy = ^.
\4.xy — X + ^y =^l. ' \xy=126.
^ (x + xy = 24. ^Q r 2^3 7/ = 218.
' \xy + y = 21. ' \x~y = 2.
4 f2y2 + y:=28. ^^ fxy=:4.
^ (15 + y = x. ^2 (x + 3y=16.
\xy = 2ij^. ' \3x^h2xyy^ = 12
Q^ (xy+Qx + 7y = 66. ^3 {frJ^ + y2='iS5.
\Sxy + 2x+5v = 70. ' lxv=3.
SIMULTANEOUS QUADRATIC EQUATIONS. 347
15/2^ + ^ = 9. .Q (xy = 4.
145. All equation containing two unknown numbers is
symmetrical when the unknown numbers can change places
without changing the equation ; q^q, 'i a? — 4: x y ■\ Z 1/ = 2 \
ic* 4 b 3^y \ 5 xij'^ + y^ = — 5 x^ ?/.
Example 1. Solve the equations :
( x^ + ,/ = 89 (1)
Uy = 40 (2)
Process. Add (1) to twice (2),
Siibtract twice (2) from (1),
Extract the square root 0! (3),
Extract the square root of (4),
x2 + 2xi/f 2/2^ 169 (3)
x22xy + y29 (4)
x + y = ± 13.
X  1/ = ± 3.
We now have to solve the four pairs
of simultaneous equations,
arf t/=13> x + y= 13 > x f y
X y= 3r xy = zy xy
r^13) x+y = 137
= 3 y xy =  3>'
There are four pairs of values, two of which are given by x = Jb 8,
'/ = ± 5, and the other two by x = ± 5, y = ± 8, in which the upper
signs are to be taken together, and the lower signs are to be taken
together.
Kotes : 1 . If the second members of two simple equations have the sign ± ,
we will have six simultaneous .simple equations to consider.
2. The above equations may be solved as in Art. 144, but the symmetrical
nwtliod is more simple.
K\.\Mi'LE 2. Solve the equations: \ „ ^~ „ _, \J.
^ lx^xy + y^2l (2)
Process. Divide (1) by (2), x + ?/ = 6 (3)
Square (3), x^ + 2xy\y^ = 36 (4)
Subtract (2) from (4), 3xy= lb, or xy = b (6)
Subtract (5) from (2). x^  2 x.V + y2 = 16.
Extract the square root, x — y = ± 4 (6)
Add (3) and (6) and divide the result by 2, x = 5, or 1,
Subtract (6) from (3) and divide the result by 2, y = 1, or 5.
348 ELEMENTS OE ALGEBRA.
Example 3. Solve the equations . 5 ^'^ + 2/^  ^  2/ = "8 (1)
^ ixy + x + y = 2^ (2)
Process. Add (1) to twice (2), x'^+2xy + y'^+x + ij = 156.
Factor, {x + yy + (x + y) — 156.
Regard a; + ^ as the unknown number, complete the square, and
solve, * a; + ?/ = 12, or 13 (3)
Subtract twice (2) from (1), factor, and transpose ^(x + y),
{X  3,)2 = 3 (X + y) (4)
From (3) and (4), {x  yY = 36, or  39.
Therefore, x  y = ± 6, ov ± y.1~39 (5)
 13 ± V39
Add (5) and (3), etc., a; = 9, or 3, and ^
Subtract (5) from (3), etc., y = 3, or 9, and ^ •
Example 4. Solve the equations : < ^ ^ ^
^ lx + y = 4 (2)
Process. Raise (2) to the fourth power,
x*\4x^y + ex^y^+ 4xy^ + y^=256 (3)
Subtract (1) from (3), etc.,
2x»y + 3x^y^+2xy^ 87 (4)
Square (2) and multiply the result by 2 xy,
2x»yh4x^y^^2xy» = 32xy (5)
Subtract (4) from (5), etc., x^y^ 32xy =  87.
Regard xy a.^ the unknown number, complete the square, and
solve, xy = 29, OT 3.
We now have the two pairs of equations to solve,
X ^y= 4\ X + y= 4)
xy=29) ' _xy = 3}
^ ar = 2 ± 5 \/^>
From the first, A "^ ^
l2^^2T5Vl.
(0; = 3, or 1.
' \y=\, or 3.
From the second, ^
When the Equations are Sjrmmetrical Combine them in
such a manner as to remove the highest powers of x and y.
SIMULTANEOUS QUADRATIC EQUATIONS. 349
Exercise 127.
Solve the followiug equations :
' \x + 7j = n. ' \x^xy + y^=2l.
^ ix^ + x + i/+i/=lS. g (2^+x^y^+y^ = 9Sl.
' \xy = 6. ' \x^ + X y \ i/ = 49.
^ (x^y^+2x+2y=50. ^Q (x^xy + y^=U.
\xy + x + y = 2d. ' \x + y=U.
g (2^ + yi = 52. j^ ra^4a^»v/2 + /=133.
\ X + y + xy = 34. ' \ o^ + x y + y^ = 19.
6. <^ 2:2 + 2r^  900 • 12. ^ a?» + 2r* ~ ^ *
l«y = 30. U + y = 8.
146. An algebraic expression is said to be honwgeneovs
when all its terms are of the same degree.
Thus, 9 x*<> f 3 X y* — 8 r* y* is homogeneous, for each term is of
the 10th degree and has ten dimensions.
Example 1. Solve the equations H « « « „ « « \^i
Process. Let y — vx, .iiifl substitute in both equations.
From (I), 6x2f 2t'«z«6t;x«= 12.
12
Therefore, ^'= 65»; + 2 »;« ^^^
From (2), 3x« + 2yx2 = 3 r^x^*  3.
Therefore, ^' = 3t;«2t;3 ^"^^
12 3
Equate (3) and (4). e5v + 2t^ = 3ra2t;3
Simplify and solve for », r = }, or  1.
350 ELEMENTS OF ALGEBRA.
Substitute y = 
12
= 4.
65 X 1 + 2(1)'
Substitute v = — ^ in (3),
12 25
"^ 65X + 2(f)23l'
.. x = i /j V31.
3/ = f^T^\V31.
Notes : 1. In finding the last values of x and y, it will be observed that ±
values of x gives respectively — and + values of y. This indicates that the
equations can be satisfied only by making y ^ — ^^ V2>1, when x^\ ^^^ V'6\ ;
and when x =. — ^^ V31, y must be + ^f VZl.
2. The sign T denotes precedence of the negative value.
When each Equation is of the Second Degree and Homo
geneous. Substitute v x tor y in botli equations.
Exercise 128.
Solve the following equations :
^ ^ x^ + xy = lb. n ( x^— o xy + y'^ = — 1.
' \y^ + xy= 10. ' \Sx^xy + Sy'^ = lS.
2 (x''xy = 24:. ^ (2x^5xy+3y'^=l.
\ X y — y^ — 8. * \ 3 x^—5 xy+'2y'^ = 4=.
= 21.
18.
2 (x^ + 4:Xy=lS3. g (x^~2xy
' \4:xy + 16/ = 228. * I xy + y^ =
^ (2x^+3xy=26. ^ (x^+3xy =
{Sy'^+2xy= 39. ' \xy + 4:y^ =
^ ( 4^2_^to/+4?/2=13. . ^ ( x^ + xy + 2 y^
' I 8x'^12xy+Sy^=ll. [2x^+2x1/ + y^
= 54.
115.
74.
■12^?/+8/=ll. " 12^24.2^^ + ^=73.
Queries. What is a homogeneous equation % Into what forms
may simultaneous quadratic equations, M'hich can be solved, be
grouped % What is the degree of the equation arising from eliminat
ing one unknown number from two equations, each of the second
degree ? Prove it. How may such equations be solved .^
SIMULTANEOUS QUADRATIC EQUATIONS. 351
Note. In solving the following equations the student is cautioned not to
work at random, but to study the equations until lie sees how they may be
combined in oixler to produce sinjple equations, and tlien perform the opera
tions thus suggested. Usually the operations of addition, subtraction, multi
plication, division, or factoring will effect a simplification of the equations.
Miscellaneous Exercise 129.
Solve the following equations :
\xy= 15.
11.
^2^+:\u.^y\:\xy^\2y^ = {).
U2r.t// + y2^1x'^/.
2 /^3' = 3
12.
{.ly\ .I25x = y — x.
I y — .0 X = .7o X y — o X.
^■\x.>,= l.
13.
f .Sx + .l2by = oxy.
1 3 a: \y = —2.25xy.
<x> + y' = 2U.
^ \x + y = 22.
14.
1 a:* f 7/4 = 706.
\x + y = 2.
(a? + i/ = 7i.
{ xy = o5.
15.
{x + y + x^\y^=\S.
1 xy = 6.
^ j(y_i)^.2_3^^2.
16.
{4(x + y)^3xy.
1 X + y \ x^\i/= 26.
1 0^7/2= 175.
17.
(a^+ ;/ ,, 337.
\x^y = 7.
(.,5+ 5 7/2 = 6 a;. ix^ + xy\x=\4.
\x^5y^ = 4:xy. ^^' \ y^ { xy + y = 2S.
^ r 2:2 4. ^y^ 140. ^^^ {2^yS = 20S.
' \y'^ + xy = 06. " I xy(x — y) =z48.
10. i^^r! 20. (?/ + =^y = 12
xyy' = 4. ■ \y + sc>y=lS.
352
21.
ELEMENTS OF ALGEBRA.
22.
^ \ if' ^=^'^xy.
X + y = b.
2xy +12 = Zx\
6 xy { 12 = a;^
2 3
23. ^ ^ y
lxy = 2.
x\ y
,./:
^4.
24.
25.
26.
27.
29.
30.
31.
32.
l\ xy
I  xy
2^4 + a;V + y/* = 7371.
x^ — xy \f = 63.
x^ + y^ = 641.
0^2/(^2 + 2/^) = 290.
x^ \ 3 a; ?/ + ;?/ = 19.
a;2 + ;^2 ^ lo;
a;2 — 3 xy + ;/^.= — 5.
3^2_5^y+3y2^9
y^ — x^ = a^.
y — X = a.
x^\y^ = Ux^7/.
X + y = a.
96 — x^y^ = 4:xy.
X + y = 6.
x^ — y/ = 56.
x^^ xy \f^ 28,
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
y'^ — xy = a? ■{■ W'.
xy — x'^ = 2 ah,
x^y — y — 21.
x^ y — X y — 6.
^ 2 x^ 480
2/2 + "^  "49" •
(^a;2 + 2/^ = 65.
1 1 a; + ?/
X y 6
a; + V 5
/
x \ y [ I
xP'y'^ + 5 a;?/ = 84.
a; + ?/ = 8.
■'« + y + Va: + y = 12.
x^ + f^ 41.
a; + ?/ + ^/x + y = 12.
x^ + y'^ = 189.
a; + Vxy + ?/ =: 19.
x^ + xy + y^ = 133.
a/^ + Vy = 4.
V^3 ^ ^f = 28.
a:?^ 4 2/5 = 6.
a:l + 7/1 = 126.
?/2 — a;2 = 4 a &.
xy = a^ — h'^.
2x? ?yxy + ?y2 = 4.
2a;2/ 3 2/2:i;2^9.
SIMULTANEOUS QUADRATIC EQUATIONS. 353
48. <i^ 2^
La; y
X ^ y xj_y _ 10
49. { x — y X + y~ 'S '
y + / = 45.
X •\ y ^ x — y
50. 'ia^b~a — b
xy = ab.
51. The sum of the squares of the digits composing a
number of two places of figures is 25, and the product of
the digits is 12. Find the number.
52. There are two numbers whose sum, multiplied by
the greater, gives 144, and whose difference, multiplied by
the less, gives 14. Find the numbers.
53. The sum of the squares of two numbers is a, and
the difference of their squares is b. Find the numbers.
Solve for a = 170 and 6 = 72.
54. A number divided by the product of its two digits
gives the quotient 2 ; and if 27 be added to the number,
the digits are reversed. Find the number.
55. The sum of two numbers is a, and the sum of their 4th
powers is b. Find the number. Solve for a = 4 and b = 82.
56. The difference of two numbers is a, and the differ
ence of their cubes is 7 a' Find the numbers.
57. The difference of two numbers is 3, and the difiTer
ence of their 5th powers is 3093. Find the numbers.
354 ELEMENTS OF ALGEBRA.
58. A number consisting of two digits has one decimal
place ; the difference of the squares of the digits is 20, and
if the digits be reversed, the sum of the two numbers is 11.
Find the number.
59. Find three numbers whose sum is 38, such that the
difference of the first and second shall exceed the difference
of the second and third by 7, and the sum of whose squares
is 634.
60. The small wheel of a bicycle makes 135 revolutions
more than the larger wheel in a distance of 260 yards ; if
the circumference of each were one foot more, the small
wheel would make 27 revolutions more than the large
wheel in a distance of 70 yards. Find the number of feet
in the circumference of each wheel.
61. Find two numbers such that their difference shall
be a, and the product of their n\k\. roots c. Solve for a = 4,
c = 2, and n = h.
62. Find a fraction such if the numerator be increased
and the denominator diminished by 2, the result will be its
reciprocal; while if the numerator be diminished and the
denominator increased by 2, the result will be \% less than
its reciprocal.
63. A principal of $10,000 amounts, with simple in
terest, to $14,200 after a certain number of years. Had
the rate of interest been 1 % higher and the time 1 year
longer, it would have amounted to $15,600. Find the
time and rate.
64. A sum of money at interest amounted at the end of
the year to $10,920. If the rate of interest had been 1 %
less, and the principal $100 more, the amount would have
been the same. Find the principal and rate of interest.
INDETERMINATE EQUATIONS. 355
CHAPTER XXIV.
INDETERMINATE EQUATIONS.
147. Simple Indeterminate Equations are equations of
the first degree that admit uf aii unlimited number of
solutions.
Thus, in 3x2y = 2, if y = 2, x = 2; if 3/ = 3, a; = 2f ; if y = 5,
X = 4; if y = Si X = 6 ; etc. It is evident that an unHmited num
ber of values may be given to y and x that will satisfy the equation.
Hence, an ecj nation containing two unknown numbers admits of as
many solutions as we please, and is indeterminate.
Since the values of the unknown numbers are dependent upon
each other, they may be confined to a particular limit ; as, for exam
ple, suppose the variables to be restricted to positive or negative inte
gers, we may thus limit the number of solutions.
Example 1. Solve 19xiby= 119, in positive integers.
Solution. Transpose 19 a:, 5i/=119— 19 2:.
Therefore, y = 233x + 4l^] (1)
lx
Since the value of y is to be integral, then  must be integral,
although fractional in form; and so also is any multiple of it.
Let — r— = n, an integer.
Therefore, x=l5n (2)
Substitute in (1), y = 20+19 n (3)
We must take only such integral values for n as will give positive
integral values for x and y.
(2) shows that n may be 0, or have any negative integral value,
but cannot have a positive integral value.
356 ELEMENTS OF ALGEBRA.
(3) shows that n may be and — 1, but cannot have a negative
integral value greater than 1 .
Therefore, n may be and — 1.
^^"^^'^=2i}'^^'%=i}'
Query. Can n be — 2 or + 1 ? Why ?
Example 2. Solve 7 x — l^y = 19, in positive integers.
Process. Transpose and solve for a;, x = 2hy + bl — ;=— J (I)
Let —j — = n, an integer.
Therefore, y = 7nl (2)
Substitute in (1), x=l2 7i\l (3)
Evidently x and y will both be positive integers if n have any
positive integral value.
Hence, x = 13, 25, 37, 49, ... .
y= 6, 13,20, 27, ....
Notes : 1. Having obtained a few of the possible values of x and y, the law
will become evident.
2. It will be seen from the above solutions that when only positive integral
values are required, the number of solutions will be limited or unlimited ac
cording as the sign connecting the terms is positive or negative.
Example 3. Solve I90x — 23y = 708, in least positive integers.
/3  x\
Process. Solve for y, y = Sx~30 — 6\—^j (1)
S — x
Let ^ = n, an mteger.
Therefore, x = 32Sn (2)
Substitute in (1), y = 6190 n (3)
Evidently x and y will both be least positive integers if n be — 1 .
Therefore, n = — ], x = 26, and y = 184.
Note 3. If the coefficient of the unknown number in the numerator of the
fraction is not 1, it will be necessary to make several transformations.
Example 4. Solve 21a:+ 17 y = 2000, in positive integers.
» ._ — ", ail xiiiegei.
3n
(2)
3»j
— 7 — = w, an integer.
n = 3  4 m.
x= 17m10
(3)
3/= 130 21 m
(4)
INDETERMINATE EQUATIONS. 357
11 4x
Solution. Solve for y, y = 117  x + — r= — (1)
w U4x
Transpose, y + a: — 117 = — yj
Since x and y are to be integral, y + x — 111 will be integral ;
11 — 4 a: .,, , ,
hence, — r^ — will be integral.
ll4a:
Let
Therefore,
>.x 3n , .
Now —z — mwtt be integral.
Let
Therefore,
Substitnte in (2),
Substitute in (1),
(3) shows that m may have any positive integral value, but can
not be 0, or have any negative integral value.
(4) shows that m may have any integral value from to 6, or any
negative integral value, but cannot have a positive integral value
greater than 6.
Therefore, m may be 1, 2, 3, 4, 5, Q, giving the following pairs of
values :
x= 7, 24, 41, 58, 75, 92.
y = 109, 88, 67, 46, 25, 4. Hence,
To Solve a Simple Indeterminate Equation, Involving Two
Unknown Numbers, for Integral Values. Find tlie value of
one of the unknown numbers. Place the fractional part of this value
efjual to Uj an integer, and solve the resulting equation for the other
unknown number. Substitute this result in the value first obtained.
Solve the two simple equations thus fonuerl, by inspection, for inte
gral values of n.
Notes: 4. It is better, in solving the original equation, to solve for the
unknown number which has the least coefficient.
5. A little insrenuity in arranging the terms will often obviate the necessity
of a second transformation.
358 ELEMENTS OF ALGEBRA.
148. There can be no integral values of x and y in an
equation of the form ax ±h y = c, ii a aiid b have a com
m6n factor not common also to c.
For, suppose d to, be any factor of a and also of b, but not of c,
sucli that a — md and h = n d.
c
Then mdx ±.ndy = c, or mx ztny = j.
Since m and n are integers, if 'x and y be also integers, mx ±. ny
is an integer. But ^ is a fraction. Hence, no integral values of
X and y can be found.
Notes : 1. If a, b, and c have a common factor, it should he removed by
division, then proceed as in Art. 147.
2. The solution of any indeterminate equation of the form ax — by =. ±c,
in which a and h are prime to each other, is always possible, and admits of an
unlimited number of integral solutions (Ex. 2, Art. 147). If the equation be
of the form ax\ hy = c, the number of results Avill always be limited ; and,
in some cases, the solution is impossible (Ex. 1, Art. 147).
Exercise 130.
Solve in positive integers :
1. 2x\3y = 2o; Ux = 5y7; 3x = 8y16,
2. 5r?:+ ll2/ = 254; 9 a: + 13 2/ = 2000.
3. 15xl7y=l; 13x9y=:l; 9xUy=:10.
Solve in least positive integers :
4 3^ + 7y = 39; 3x+4y=:39; 7x+lDy = 22^.
5. 27a;192/ = 43; 2^ + 7?/ =125; 555y22a: = 73.
6. 19 ^5?/ =119; I7x = 4:9yS.
Are integral solutions possible for the following? Why?
7. 3^ + 21 7/ = 1000; 7 a; +14?/ = 71.
8. 323 a? 527 y= 1000; 166 a:  192 y = 91.
INDETERMINATE EQUATIONS. 369
9. Solve 7 a 4 15 y = 145, in positive integers, so that
X may be a multiple ot y.
^ 146 , 145 n
Suggestion. Let xny, then y = ^ ^^ , and x = ^^^^^ .
10. Solve 'S9 X — 6 y = 12, in positive integers, so that
y may be a multiple of z,
11. Solve 20 a: — 31 2/ = 7, so that x and y may be
positive, and their sum an integer.
Suggestion. Put x + y = n.
149. A problem is indeterminate when it involves less
conditions than there are unknown numbers.
Exercise 131.
1. Find a number which being divided by 3, 4, and 5,
gives the remainders 2, 3, and 4, respectively.
Solution. Let x represent the number and y the sum of the
quotients, then,
x2 a3 x4
3 +^ + — = »•
f'S\y\
Simplify and solve for ar, a = y + 2 4 13 I — j;^ I (1)
3lw \ •*' /
Let ^= = n, an integer.
Therefore, y = 47 n — 3.
Substitute in (1). a: = 60 n  1.
Hence, n may be 1, 2, 3, 4, etc.
Therefore, x = 59, 119, 179, 239, etc.
.7 = 44, 91, 138, 185, etc.
2. Find the least number which being divided by 2, 3,
4, 5, and G, gives remaindeis 1, 2, 3, 4, and 5, respectively.
3. Find two numbers which, multiplied respectively by
14 and 18, have for the sum of their products 200.
360 ELEMENTS OF ALGEBRA.
4. Divide 142 into two parts, one of which is divisible
by 9, and the other by 14.
5. There are two unequal rods, one 5 feet long and the
other 7. How many of each can be taken to make up a
length of 123 feet ?
6. Find two fractions having 5 and 7 for denominators,
and whose sum is .
7. Find the least number that when divided by 9 and
17 will give remainders 5 and 12, respectively.
N 5
Suggestion. Let N represent the number, —  — = x, and
N~ 12
— p^ — = y. .'. 9x=l7y{1.
8. A farmer bought sheep, pigs, and hens. The whole
number bought is 125, and the whole price, $225. The
sheep cost $5, the pigs $2.50, and the hens 25 cents.
How many of each did he buy ?
Solution. Let
X = the number of sheep,
y = the number of pigs,
and
z = the number of hens.
Then,
x +
y + z= 126 (1)
and
5x+2.
■5y +
.25 z = 225 (2)
From (1) and (2),
?/=862x^ (3)
Let
x1
Q  n, an integer.
Therefore,
x = 9n\l.
Substitute in (3),
y = Ml9n.
Substitute in (1),
2;= 40+ 10 n.
Therefore, n may be 0, 1, 2, 3, and 4, giving the following values :
x= I, 10, 19,28,37.
y = 84, 65, 46, 27, 8.
z = 40, 50, 60, 70, 80.
PROBLEMS. 361
Qaeries. How many solutions ? In how many diflFerent ways
may the stock l)e bought? How solve by means of only two un
known numbers?
9. How can one pay a sum of $ 1.50 with 3 and 5 cent
pieces ? In how many ways can the sum be paid ?
10. Can a grocer put up the worth of S3.50 in 11 and
7 cent sugar ? In how many ways can he do it in even
and odd pounds, respectively ? Find the greatest and least
number of pounds of the 7 cent sugar he can use.
11. Is it possible to pay £50 by means of guineas and
threeshilling pieces only ?
12. A owes B $5.15. A has only 50cent pieces and B
only 3cent pieces. How may they settle the account ?
13. A farmer bought horses at S 60 a head and sheep at
S8, and found that he bad invested S4 more in sheep than
horses. How many of each kind did be buy ?
14. A farmer invested $1000 in 75 head of cattle, worth
$25, $15, and $10 per head. Find the number of each
kind, and the number of ways in wliich he could buy
them.
15. A grocer had an order for 75 pounds of t€a at 55
cents a pound, but having none at that price he mixed
some at 30 cents, some at 45 cents, and some at 80
cents. How much of each kind did he use, and in how
many ways can he mix it?
16. How many pounds of 20, 35, and 40 cent coffee
must a grocer take to make a mixture of 150 pounds worth
30 cents a pound ? In how many ways can the mixture
be made ?
362 ELEMENTS OF ALGEBRA.
17. How many gallons of S 1.50, S1.90, and $1.20 wine
must a vintner take to make a mixture of 40 gallons worth
$1.60 per gallon ? How many ways may the mixture be
made ? Can an odd number of gallons of each kind be
taken ? An even number ?
18. In how many ways can £1 be paid in halfcrowns,
shillings, and sixpence, the number of coins in each pay
ment being 18 ?
19. A hardware merchant paid $180 for 20 stoves.
There were three sizes: one $19 each, another $7, the
other $6. How many of each size did he buy?
20. A person having a basket of oranges, containing
between 50 and 72, takes them out 4 at a time, and finds 1
over; he then takes them out 3 at a time, and finds none
over. How many had he ?
A^l N
Suggestion. Let N represent the number, — j— = x, and « = t/.
l+x ^ I + x
.'. y = x+ ~^' Pwt — ^ = n. Then n must be 5 or 6.
21. A poultry dealer has a basket containing between
200 and 300 eggs, he finds that when he sells them 13 at a
time there are 9 over, but when he sells them 17 at a time
there are 14 over. Find the number of egfjs.
22. Two countrymen together have 100 eggs. If the
first counts his by eights and the second his by tens, there
is a surplus of 7 in each case. How many eggs has each ?
23. A surveyor has three ranging poles of lengths 7 feet,
10 feet, and 12 feet. How may he take 40 of tliem to
measure 113 yards? In how many ways may the mea
surement be made ?
INEQUALITIES. 363
CHAPTER XXV.
INEQUALITIES.
150. Since a positive number is greater than any negative num
l)er, the statement that a is algebraically greater than 6, or that a—h
is positive, is expressed by a > 6 ; that a is algebraically less than 6,
or that a — 6 is negative, is expressed by a < 6. Hence,
An Inequality is a statement that on^ expression is
greater or less than another; as,
1 — ar > = — : m — n < x.
The expression at the left of the sign is calle<l the first member,
and the expression at the right, the second member of the in
equality.
The fonn a> h> c, means that 6 is less than a but greater
than c.
Notes : 1. Inequalities are said to subsist in the same sense wlien the lirst
member is the greater in each, or the first member is the less in each ; as,
3 > 2, 7 > 5, and .5 > 3 ; a<h, c<d, and m<n.
2. Two inequalities are said to subsist in a contrary sense when the first
member is the greater in one, and the less in the other ; as, 5 > 3 and o < 6 ;
m < 5 and h > n.
3. An inequality is said to be solved when the limit to the value of the
unknown numl)er is found.
151. Subtract a + & from each member of a > 6,
then, a  (a + b) > b (a \ 6).
Simplify, . — 6 > — a,
or, ~ a < — b Hence,
I. If each member of an inequality has its sign changed, the sign of
inequality will be reversed.
364 ELEMENTS OF ALGEBRA.
Multiply each member of
 5 < 5 by 
2,
then,
10 >  10.
Multiply each member of
a > 6 by 
m,
then,
— am < — 6m.
Divide each member of
 6 < 4 by 
2,
then,
3 >  2.
Divide each member of
a > ft by 
m,
then.
a b
m m'
He
II. If each member of an inequality be multiplied or divided by
the same negative number^ the inequality tvill be reversed.
Suppose a> b, c > d, m'> n, —
By definition, a — b, cd, mn, are positive.
Add, (^a b) + {c  d) + (m  n) + .... is positive.
or, (a + c + m I .)  (b + d \ n \ ... .) is positive.
Therefore (by definition), a + c ^ m + — > b + d + n + .,»,
Thus, 7 > 3
5> 2
4> 1
Add, 16 > 6, or divide by 2, 8 > 3. Hence,
III. If the corresponding members of several inequalities be added,
the sum of the greater members will exceed the sum of the lesser
members.
Suppose a > b and m > n, then a  b and m — n are positive.
But, (a — b) — (m — n), or (a — m) — (6 — n) may be either
positive, negative, or 0.
Therefore, a — m > b — n, a — m < b — n, or a — m = b — n.
Thus, 5 > 3
3> 2
Subtract, 2 > 1
7>4
5> 1
Subtract, 2 < 3
8> 7
6>5
Subtract, 2 = 2. Hence,
IV. If the members of one inequality be subtracted from the corres
ponding members of another, tht resulting inequality will not always
subsist in the same sense.
INEQUALITIES. 865
1 2 X 3 z 64
Example 1. Solve 3^ x ^ — > ^^ + ^^ for the limits of x.
Solution. Free from fractions and simplify,
112a;6>45x+ 128.
Subtract 45 x  6, 67 a; > 134.
Divide by 67, x > 2.
Therefore, x is greater than 2.
Example. 2. Solve the following :
\bxZy>Zx^b (1)
i3a: + t/=r22 (2)
Solution. Subtract 3 x from (1), 2 x  3 y > 5 (3)
Multiply (2) by 3, 9x + 3 y = 66 (4)
Add (3) and (4), 11 x > 71.
Divide by 11, x > Q^^.
22 w
From (4), x = ^ •
Substitute in (3) and simplify, —y>\\.
Therefore, 2/ < 2^^ (see I)
Example 3. Solve the inequalities :
{fix — mn^n^ — mx (1)
\ mx ~ nx + mn < in^ (2)
Process. Simplify (1) and solve, x > n.
Simplify (2) and solve, x <rr.
Hence, x is greater than n and less than m.
Note. The principles applied to the solutions of equations may be applied
to inequalities, except that if each member of an inequality haa iti sign
changed, the lign of inequality will be reversed.
Exercise 132.
Find the limit of x in the following :
1. 4x3 >fa;f; f —  ^ < 9  3 a:.
^o o o 1 x^ — a a — X 2x a
366 ELEMENTS OF ALGEBRA.
o ax — 2b a X — a ax 2
^ Zh 2h~ ^ 1" ~ 3 *
4. If 2^2 + 4 a; > 12, show that x>2.
5. If 7 a;2  3 a; < 160, show that x < o.
6. If 4 ic + 12 — ic^ > 0, show that x is included be
tween 6 and — 2.
7. If 9 a; < 20 it^ + 1, show that x > \ oi < \.
8. If 15 — ic — 2 ic^ > 0, show that x lies between 
and — 3.
Q I ^ ;^ > 30 — 4 ic. ^Q ( 1 J ^ < I ^ + 3.
< 3 :?: + 49. * ( 6 oj > 24  2 ^.
Find an integral value of x in the following
\ IQx
( (^ + 2) + ^ < I (^  4) + 9.
' \l(x\2) + \x>l{x+l)^^.
Find the limits of x and y in the following :
.o i3^45^>121. 7.^; + 5^>19.
* (4a;+ 72/ =168. / a;  3/ = 1.
(« + &) ^ — (a — ?)) 2/ > 4 a &.
^ I (tt  &) ^ + (a 4
h)y = 2(a + h) {a  6).
15. A certain number plus 5, is greater than one third
the number plus 55 ; while its half plus 2, is less than 41.
Find the number.
INEQUALITIES. 367
16. Find the price of oranges per dozen, when three
times the price of one orange, decreased by tiiree cents, is
more than twice its price increased by one cent ; and eight
times the price of one orange, decreased by twenty cents;
is less than three times its price increased by ten cents.
152. Since the stjuare of a negative number is positive, if a and b
represent any two numbers, (a — 6)^ must be positive, whatever the
values of a, and b. Therefore, since every positive number is greater
than zero,
(a  6)2 > 0.
Expand, a^  ^ah + h'^> 0.
Add 2 a 6 to each member, a* + 6^ > 2 a 6. Hence,
The sum of the squares of two unequal numbers is greater
than tvnce their product.
Vote. The above is a fundamental principle in inequalities.
Example 1. Show that a"^ \ 1)^ ■\ c^ > ab + a c + h c, a and h
positive.
Prool Since a, b, and c are any unequal numbers,
a2+62>2a6 (1)
a«+c«>2ac (2)
62 + c«>26c (3)
Add the corresponding members of (1), (2), and (3),
2aa + 2 62l2c2>2a6+2ac + 2 6c.
Divide by 2, a« + 6« + c^ > a 6 + a c + 6c.
Query. How if a = 6 = c ?
Example 2. Show that a« + 6« > a* 6 } a 6*.
Proof. We shall have, a* + }fi > a'^b + a 6*.
Factor, (a + 6) (a«  a 6 + 6^) > a 6 (a + 6).
Divide by a + 6, a'^ — ah \ b"^ > ab.
Add a 6, a« + 6^ > 2 a 6.
Therefore, a« + 6» > a^ 6 + a 6*.
368 ELEMENTS OF ALGEBRA.
Example 3. Which is the greater, V/ 1 1/ — r or /y/a i
Proof. We shall have,
Square each member,
a2 62
2 Y a bmn \ r > or < a 6 + 2 \/a bmn + m n.
it9f tV CI O
, a^h^ m^n^
Subtract 2 \/ahmn^ 1 — r > or < ao + ww.
^ m n a
Free from fractions and factor,
{ah + m n) {a^ h^ — abmn \ m^ n^) > or < abmn(ah + mn).
Divide by ab \ mn,
a^h^ — abmn + m^n^ > or < ab mn.
Add a b 7« n, a^ b^ + m^ n^ > or < 2 a 6 m n.
But, a^b^i m'^n^':> 2abmn.
Therefore, \ 1^ + \ afT > ^""^ ^ V^""
Exercise 133.
Show that, the letters being unequal, positive, and
integral :
h^ a^ a b
2. a 62 + a% > a^+ b^ ; (ni^] n^) (mH n^) > (m^+ n^f.
3. xy\xz\yz < {x + y^z)^\{x+z—yf'{{y + z~x)\
Which is the greater :
. „ . „ , a + b 2 ab m n 11
4. 71^41 or ?i2+ ^ ; —^ or —7 ; 5 s or
5. 1±J or ^^; 3(1 + a2 + a^) or (1 + c^ + a'f,
ga^ y x^ — y^ ^
INEQUALITIES. 369
1/9 a/^
V3 v5
Queries. How in 4 and 6, if a = 6 ? In 4, if n = 1 ?
7. If a^ + ^>2 + c2 = 1, and a?» + y» + «2 = 1, show that
aa; + &y + C2 < 1.
Query. How ifa = 6 = c = a: = y = z?
If a > 6, show that :
8. a  6 > iVa  V6)^ a^ + 7 aH > (rr + &)»
9. a6* > a*6« ; a^ + 13 a ^2 > 5 a26 + U 63.
Miscellaneous Exercise 134.
Example 1. Solve the inequahties :
{:
V2(xyfy2) + 4<;y(2i/l)(y + a:) (1)
2x + 5y>8 (2)
Solution. Square each member of (I) and simplify,
2xy\2y^\ 4 <2y^^2xy yx.
Subtract 2ari/ + 2 y«, 4 <yx (3)
Multiply each member of (3) by 2,
8<2y2a: (4)
Add the corresponding members of (2) and (4),
3y > 16. .. y >5f
Multiply each member of (3) by 5,
20<5y5a: (5)
Add (2) and (5),  3a: > 28, or 3a: < 28. .. x < 9f
Example 2. Simplify (y\x<m — n) (m* + m n 4 71^ > y — x).
Solution. We are to multiply the corresponding members
together, (y + x) (y — x) = if — a:*,
(m — n) (m* ■\ mn \ n'*) = m* — n*.
Therefore, (y +x < mn){mHmn\ n^ >yx) = y^~x^ < m* n».
24
370 ELEMENTS OF ALGEBRA.
Example 3. Which is the greater, x^ + y^ or x^y \ y*xl
Proof. We shall have, x^ + y^> or <: x* y { y* x.
Subtract x^y \ y^x, x^ — x^ y + y^ — y* x > or < 0.
Factor, (x^  y^) (xy)> or < 0.
Now, whether a; > or < .y, the two factors, x^ — y^ and x — y, will
have the same sign. Hence, since (x'^—y^) i^~y) is always positive,
{x^  y^) {xy)> 0.
Therefore, a;^ + 2/^ > x^ y \ y^x.
Example 4. Which is the greater, m* — n* or 4 n^ (^ _ ,^) when
m > n ?
Proof. We shall have, m^ — n* > or < Am\m—n).
Divide by w — n, m^ + wi*n + m w^ + n^ > or < 4 m^.
Subtract wi^ + m^ n and factor the resulting inequality,
n^ (m + n) > or < m\^m—n).
But, m > w (1)
Square (1), m^ > n^ (2)
Multiply (1) by 2, 2 m > 2 n.
Add m — n, 3m — ??>m + n (3)
Multiply the corresponding members of (2) and (3),
m^ (3 m — n) > n' (m { n).
Therefore, 4m^ (m — n) > w* — n*.
6. Find the sura of x^ + y > 1 — a, y^ — 2 a > 5 + 4,
^ X + y < 2 a + 1, and y^  S x^ < 5  a.
6. From a^ + 2 a a;^ < 5 take a (a + x^) y n^ —1.
7. From a2 < 3  .7.2 subtract 2 ^2 > 5.
Multiply :
8. {a + hf > {xyf by 3; Zf < 5^^ by x^ + y\
9. Divide a^l^ > a^+ h^ by a2 + ^2
10. Divide 11 a2 + 88& > 121^2 ^y  H.
INEQUALITIES. 371
Perform the indicated operations and simplify :
11. (wl< 5)(m + l<10); (a < n + h){nb > c).
12. (_ 2 >  3)3 ; (5 > 2) f (3 < 4) ; V25 > 9.
13. [ 243 >  1024]i ; (71 + 1)^ > n^  n^ + 4: n.
14. m3  7i3 > (m  n) {m^ + n^) ; 4^ 64 < 8.
15. (m2  n^ < u.^)'T{ix> m + n); [ ?t > i/f.
Solve :
16. {X''2f > 0^+ 6 x25',V(xl)^ + 'S 2^ + 6 >2xK
17. a;  2 > V ^^=^ ; V3  4 Vi > VI6 2:  5.
^g 3y+2a;>3. ^^  a: + f > Vo^  3 a; + y.
( 4 > 4 7/ + ic. ■ ( 5 > a: — y.
j42/a^>?/ + 4. J3:rl>a: + 3y.
^" (3a;6y> l4i/. "^ ( 27/3rr2 = 3a;3.x2.
22. 38a:715a:2<0; 6 ar^ 4 7 ;r + 2 < 0.
23. 17 a; 6r»5<0; 6 a: + 11  a:^ < 2 a;  10.
Find integral values of x in the following :
(3ix.5x>5. {x + 7
^^ l2.5a: + ia;<8. ^^' (2a; +
a; + 7 < 15.
10 > 20.
26 U^i^<3. 27 2a:5>31.
I 7a;15>4a;+30. ^' (3a;20<2ar.
28. ar2 4 2a;15<0; a:^ ^ lOa: 4. 63 < 0.
372 ELEMENTS OF ALGEBRA.
Show that:
29. Vl9+V3> VIO + VT; V5 + Vn > V3+3V2c
li a >h, show that :
30. Va^  62 + Va^  (a  hf > a.
31. a^h^ <3a^{a h) and >Sb^{a b).
32. ah> j^ and < ^3 .
If x^ — a^ + y^, if' — (p ^ d'^^ show that:
33. xy'^ac + bdoTad + bc.
Show that:
34:. {a b + X I/) (a X + b y) ^ 4: a b xy,
35. (ti + 6) (ft + c) (6 + c) > 8 a & c.
36. Show that the sum of any fraction and its reciprocal
is greater than 2.
?7. In how many ways may a street 20 yards long and
15 wide be paved with two kinds of stones ; one kind
being 3f feet long and 3 wide, the other 4 feet long and
4 wide ?
38. A and B set out at the same time to meet each
other; on meeting it appeared that A had travelled a miles
more than B, and that A could have gone B's distance
in n hours, and B could have gone A's distance in m
hours. Find the distance between the two places. Solve
when ft = 18, ^ = 378, and m = 672.
SERIES. 373
CHAPTER XXVI.
SERIES.
153. A Series is an expression in which the successive
terms are formed according to some fixed law ;
As, 1, 2, 4, 8, , in which each term is double the preceding
term ; a, a + c/, a + 2d, a + 3d, , in which each term exceeds
the preceding term by d.
ARITHMETICAL PROGRESSION.
154. The expressions 1, 6, 9, 13, 17, , and 16, 10, 6, 0,
— 5, —10, ...., are called arithmetical progressions or series. The
first is an increasing series, and the second a decreasing series.
The general form for such a series is,
a, a + d, a\^d, a + 3<i, a + 4ci, a + Hd, a + 6cl, ....
in which a is the first term and d the common difference ; the series
will be increasing or decreasing according as d is positive or negative.
Hence,
An Arithmetical Progression is a series in which the adja
cent terms increase or decrease by a common difference.
In every arithmetical series the following elements occur, any
three of which being given, the other two may be found :
The first term, or a.
The last term, or /.
The common difference, or d.
The number of terms, or n.
The sum of the terms, or s.
374 ELEMENTS OF ALGEBRA.
By an examination of the general form it is seen that the coefficient
of d is always 1 less than the number of the term.
Thus, the 2d term is « + rf, or a + (2 — 1) cf,
3d term is a + 2 </, or a + (3 — 1) rf,
4th term is a f 3 </, or a + (4 — 1) 6?,
12th term is a + 11 rf, or a + (12 — 1) d, and so on.
In the nth, or last term, the coefficient of c? is n — 1. Hence,
To Find the Last Term of an Arithmetical Series, when the
first term, the common difference, and the number of terms
are given.
I = a+ {nl)d (!)
Note. The common difference may always be found by subtracting any
term of the series from that which immediately follows it.
Example 1. Find the 18th term of the series , f, , etc.
Process. Here, n = 18, a = , and d — ^ — l = \.
Substitute these values in (i), / =  + (18  1) J = 4.
Example 2. Find the 30th term of the series x \ y, x, x — y, etc.
Process. Here, n = 30, </ = x — {x+y) = —y, and a = x + y.
Substitute these values in (i), I = x + y + (30 — l)(—y) = x — 28y.
Exercise 135.
Find:
1. The 15th term of 7, 3,  1, ....
2. The 27th and 41st terms of 5, 11, 17, ....
3. The 20th and 13th terms of  3,  2,  1, ....
4. The 37th and 89th terms of 2.8, 0, 2.8, ....
5. The 40th term of 2 a — h, 4 a  3h, 6 a — 5h, ..
6. The 15th and 8th terms of J, 1 . 
ARITHMETICAL PROGRESSION. 375
7. The first term is J, the 102d is 18. Find the com
mon difference.
8. The 21st term is 53, and the common difference is
— 2J. Find the first term.
9. The first term is 5^, and the common difference is 3 J.
What term will be 42 ?
10. The first term is ^, the common difference is , and
the last term is 17^. Find the number of terms.
11. The 54th and 4th terms are  125 and 0. Find
the 42d term.
12. Find three terms whose common difference is J,
such that the product of the second and third exceeds that
of the first and second by. 1 J.
155. Taking the elements as given in Art. 154 :
s = a + (a + rf)+(« + 2</) + (a + 3rf)+(a + 4f/) + ••.. ^
or 5 = / + (;r/)f (/2c?H(/3£/) + (/4cO + ... a
Add, 2s = {a + [)\{a + l)\{a\l) + (a + l) + (a + l) f .... to n terms,
or 25 = n(a + /) (1)
Substitute the value of I from (i) (Art 164) in (I),
a » = n [2 a + (n — 1) d]. Hence (solve for s),
To Find the Sum of all the Terms of an Arithmetical Seriei
» = ^[aaf (nl)<f] (iH)
Example 1. Find the sum of an arithmetical series of 17 terms,
the first term being 5^, and the last term 25 i.
Process. Here, n = 17, a = 5^, and / = 2.''>.
Substitute these values in (ii), « = y. (5^ + 25^) = 263j.
376 ELEMENTS OF ALGEBRA.
Example 2. Find the sum of the series 3, 1,1^, , to 19
terms.
Process. Here, n = 19, a = 3^, and c? = 1 — 3^ = — 2^.
Substitute these values in (iii),
s= ^[2 X 3^ + (19  1)( 2^)] =  361.
12 3
Example 3. Find the sum of m , 3 m , 5 m , . . . . , to
m' m m' '
m terms.
Process. Here', n = m, a = m — — , and d =3m —~ ~ m — )
= 2m —  •
m
Substitute in (iii),
2 m^ — m — 1
=fK'"9+^'" '>("" 9]
Example 4. The first term of a series is 3 m, the last — 35 m, and
the sum — 320 7W. Find the number of terms and the common
di Herence.
Process. Here, s =i — 320 m, a = 3 m, and / = — 35 m.
Substitute in (ii),
 320 m = ■x(S m  35 m) =  16 mn. .. n = 20.
Substitute in (iii),
 320 m = %0 [6 m + 19 rf] = 60 m + 190 c?. .. d = 2m.
Example 5. How many terms of the series — C, — 6f , —6, ,
must be taken to make  52 ?
Process. Here, s= — 52, a = — 6, and c/ = f .
n
Substitute in (iii),  52 = ^ C" ¥ + (n  1) X f ].
Simplify and solve for n , n = 1 1 or 24.
Query. Do both of these values satisfy the conditions ? In
explanation write out 24 terms of the series and observe that the
last 13 terms destroy each other.
ARITHMETICAL PROGRESSION. 377
Exercise 136.
Find the sum of :
1. 5, 9, 13, ..... to 19 terms.
2. 10 J, 9, 7 J, ...., to 94 terms.
3. 3 a, a, — a, . . . . , to a terms.
4. 3 J, 2 J, 1, ...., to n terms.
^w— Im — 2 m — 3
5. , , , ...., to m terms.
m mm
. 2ag~l ^ 3 6 g^  5
6. , 4a , , ..... to n terms.
a a a
^ 4a + & 5a + 2&
7. a, — X — , ^ , ...., to 19 terms.
8. The first term is 3^, and the sum of 14 terms is 84J.
Find the last term.
9. The sum of 40 terms is 0, and the common difference
is — ^. Find the first term.
10. Find the number of terms and common difference:
(1) when the sum is 24, the first term 9, and tlie last —6;
(2) the sum 49 a, the first term a, and the last 13 a.
11. The sum of 12 terms is 150, and the first is 5J.
Determine the series.
12. Show that the sum of the first n odd numbers is r?,
13. Find the sum of all the odd numbers between 100
and 200.
378 ELEMENTS OF ALGEBRA.
14. The sum of five terms is 15, and the difference of
the squares of the extremes is 96. Find the terms.
15. Find the sum of =i, tj , 7=, ....,
l'\ Vx ^^ 1 Vx
to n terms.
156. a is called the arithmetical mean between a — d and a + d.
Hence,
An Arithmetical Mean is the middle term of three num
bers in arithmetical series.
If a and 6 represent two numbers, and A their arithmetical mean,
the common difference is A — a, or b — A. Therefore,
A — a — h — A. Hence (solve for J.),
To Find the Arithmetical Mean Between two Terms.
A = ^ M
If a and I represent any two numbers, and m the number of means
between them, the whole number of terms is m + 2, or wi + 2 = n.
Substitute this value for n in (i) (Art. 154),
I = a + (m + I) d. Hence (solve for d),
To Insert any Number of Arithmetical Means Between two
'^«" ^ . 1 (V)
This finds d, and the m required means are,
a\ d, a + 2d, a + ^d, a + 4Ld, ...., a\md.
Example 1. Find the arithmetical mean between: (1) 27 and
— 5 ; (2) rri^ \ mn — n^ and m^ — m n \ n^.
Process. (1) Here, a = 27, 6 = — 5.
275
Substitute in (iv), A = — x — = 11.
(2) Here, a — m^ + mn n^, b =m^ — mn + rfi.
^ , . . ,. ^ , m^ + mn  n^ + m^ — mn + n^
Substitute in (iv), A = 2 — =; m\
GEOMETlilCAL TUOGRESSION. 379
Example 2. Insert five arithmetical means between 12 and 20.
Process. Here, a = 12, / = 20, and m = 5.
20 12
Substitute in (v), d = ^ ^ = H
The series is 12, 13^, 14, 16, 17^, 18f, 20.
Exercise 137.
Insert :
1. 14 arithmetical means between — 7J and — 2^.
2. 16 arithmetical means between 7.2 and — 6.4
3. 10 arithmetical means between 5 m— 6 n and 5w— 6 m.
4. 4 arithmetical means between — 1 and — 7.
5. X arithmetical means between a^ and 1.
« ^. , , . , . , , m— n , mhn
6. Find the arithmetical mean between — —  and .
m+n m — n
7. The arithmetical mean between two numbers is — 9,
and the mean between four times the first and twelve times
the second is — 66. Find the numbers.
GEOMETRICAL PROGRESSION.
157. The expressions 3, 9, 27, 81, ...., and 1, i, i, ^, ....,
are called geometrical progressions or series. The general form for
such a series is,
a, ar, ar^, ar*, ar*, ai*, an*, ar'', ....,
In which a is the first term, and r a constant factor or ratio. Hence,
A Geometrical Progression is a series in which the adja
cent terms increase or decrease by a constant factor.
The Common Ratio is the fiictor by which each term is
multiplied to form the next one.
380 ELEMEJ^TS OF ALGEBRA.
In every geometrical series the following elements occur; any
three of which being given, the other two may be found.
The first term, or a.
The last term, or I.
The common ratio, or r.
The number of terms, or n.
The sum of the terms, or s.
By an examination of the general form it is seen that the expo
nent of r is always 1 less than the number of the term.
Thus, the 2d term is a r,
3d term is a r^,
4th term is a r*,
12th term is ar", and so on.
In the nth, or last term, the exponent of r is w — 1. Hence,
To Find the Last Term of a Geometrical Series, when the
first term^ the comtnon ratio, and the number of terms are
given.
I = ar""^ (i)
Notes: 1. The common ratio is found by dividing any term by that which
immediately precedes it.
2. A geometrical series is said to be increasing or decreasing, according as
the common ratio is greater than 1, or less than 1.
3. An arithmetical series is formed by repeated addition or subtraction; a
geometrical series by repeated multiplication.
Example 1. Find the 8th term of the series .008, .04, .2, etc.
Process. Here, a = .008, n = 8, and r = .04 ^ .008 = 5.
Substitute in (i), / = .008 X 58i = 625.
Example 2. Find the 10th term of — , x, y, — ,
Process. Here, a = — , n = 10, and r = x . — =  •
y\ ' y X
Substitute in (i), / =  fj = ar'yS
GEOMETRICAL PROGRESSION. 381
Exercise 138.
Find:
1. The 5th and 8th terms of 3, 6, 12, ....
2. The 10th and 16th terms of 256, 128, 64, ....
3. The 8th and 12th terms of 81,  27, 9, ....
4 The 14th and 7th terms of gJ^, ^^, 3^, ....
. rr.1 « 1 ^ X mx m^x
5. The 6th term of , — 5, — 3
y / f
6. The mth term of x, x^^ 7^, ....
7. The 3d and 6th terms are f^ and — . Find the
series and the 12th term.
8. The 5th and 9th terms are f J and §. Find the
series.
9. If from a line a inches in length, one third be cut
off, then one third of the remainder, and so on; what part
of it will remain when this has been done 5 times ? When
/ times.
168. Taking the elements as given in Art. 157,
8z=a + ar\ar^ + ar*\ \ai*^\ar^'^ (1)
Multiply (1) by r,
sr = ariar^iaH» f a r"'^ I g r»^ t g r (2)
Subtract (1) from (2),
sr—8 = ar^ — a (3)
Substitute the value of or* from (i) (Art. 167) in (3), and factor
the result, « (r  1) = r /  a. Hence (solve for <),
To Find the Sum of all the Terms of a Geometrical Series
8 = ^^null (ii)
r  1
382 ELEMENTS OF ALGEBKA.
Example 1. Find the 6th term and the sum of — J, ^,  f , .
Process. Here, a = —^, 7i = 6, and r = — .
Substitute in (i) (Art. 157), ^ =  ^ x ( )^ = ^.
Substitute in (iii), 5 = " '^_^s _ ^ ^ = W
EXxiMPLE 2. Find the least term and the sum of 3, — 9, 27,
to 7 terms.
Process. Here, a = 3, n — 1, and r — — 3.
Substitute in (i) (x\rt. 157), Z = 3 ( 3)« = 2187.
3^:^y— =1641.
Substitute in (ii),
^= 3
Exercise 139.
Find the sum of:
1. 3, 1, J,....
, to 6 terms.
2. lhi>'
,..., to 6 terms.
3. 1, J, A. ••
. ., to 8 terms.
1 3
*• V3' ' V3'
...., to 8 terms.
5. 1, 3, 32, .....
to m terms.
6. 2, 4, 8, ...
. , to 2 m terms.
7. The 7th and 4th terms are 625 and — 5. Find the
1st term, and the sum of the 4th to the 7th terms inclusive.
8. The sum of the first 10 terms is equal to 33 times
the sum of the first 5 terms. Find the common ratio.
9. The sum of three numbers in geometrical progression
is 216, and the first term is 5. Find the common ratio
and the numbers.
GEOMETRICAL PliOGRESSION. 383
159. A Geometrical Mean is the middle term of three
numbers in geometrical series.
If a and b represent two numbers, and G their geometrical mean,
G b
the common ratio is — , or ^. Therefore,
G b
— = jy. Hence (solve for G),
To Find the Geometrical Mean Between two Terms
O = \/ab ' (iv)
If a and b represent any two numbers, and m the number of means
between them, the whole number of terms is m + 2, or m f 2 = n.
Substitute this value for n in (i) (Art. 167),
/ = a r"* + 1. Hence (solve for r),
To Insert any Number of Geometrical Means Between two
Terms. , , _
This finds r, and the m required means are,
ar, ar^, ar^, ar^, ar^ , ar»*.
Example 1. Find the geometrical mean between : (1) — = and
3 V3
7^; (2) 3x»y and I2xfz.
"^^ 1 3
Process. (1) Here, a = —p, and b = —z •
Substitute in (iv), G = v/ — p ^ ^=^
3^
V3 V3
(2) Here, a = 3x*y and b = Uxy'z.
Substitute in (iv), G = ^33*y X I2xy*z = 6x^y^ ^z.
Example 2. Insert six geometrical means between 14 and  ^y.
Process. Here, a = 14, / =  /y, and m = 6.
Substitute in (v), r = ^—^\^ = — i
Hence, the series is 14, 7, f  }, , ^^h ^.
384 ELEMENTS OF ALGEBRA.
Exercise 140.
Find the geometrical mean between :
1. 7 and 252; a^h and ah^, f and i ;  and .
2. yV '^"cl jJ^o ; 4^:2  12a; + 9 and ^x^\12x^ 4.
Insert :
3. 2 geometrical means between 5 and 320.
4. 2 geometrical means between 1 and \.
5. 3 geometrical means between 100 and 2J.
6. 6 geometrical means between 14 and — ^^.
7. 7 geometrical means between 2 and 13,122.
8. Which is the greater, and how much greater, the
arithmetical or geometrical mean between 1 and \.
9. Find two numbers whose sum is 10, and whose geo
metrical mean is 4.
HARMONICAL PROGRESSION.
160. The expressions h ^, \, \, ...., and 4, f,  f, 4, ....,
are called harmonical progressions or series, because their reciprocals
1, 3, 5, 7, , and ^, — , — , — ., . form arithmetical series.
The general form for such a series is,
I ^> Wh^ ^TT^^ry. Hence,
An Harmonical Progression is a series the reciprocals of
whose terms form an arithmetical series.
HARMONICAL PROGRESSION. 385
Notes: 1. Evidently all questions relating to harmonical pro^^ression are
readily solved by writing the reciprocals of the temis so as to form an arith
metical series.
2. There is no general formula for finding the sum of the terras of a har
monical series.
3. The term harmonical is derived from the fact that musical strings of
•qual thickness and teiisiou produce harvumy when sounded together, if their
lengths are represented by the reciprocals of the series of natural numbers; that
is, by the series 1, J, J, i, J, J, etc. Harmonical properties are also interesting
because of their importance in geometry.
Example 1. Find the mth term of the series 3, IJ, 1, f , f , etc.
Solution. Taking the reciprocals of the terms, we have ^, , 1,
, , etc. ; an arithmetical series.
Here, a = , rf = J, and n = m.
Substituting in (i) (Art. 154), d = ^ + {m  I) ^ = ^. Taking
the reciprocal of this value for the required term, we have — .
Example 2. The 12th term is , and the 19th term is ^. Find
the series.
Process. The 12th and 19th terms of the corresponding arith
metical series are 5 and ^.
From (i) (Art. 154), 5 = o + 11 rf,
^ = a + 18 d.
Solving for a and rf, a =  and rf = J.
The arithmetical series is, , , 2, t, , 3, Y> • • • •
The harmonical series is, , , ^, ^, , J, i%,
161. A Harmonical Mean is the middle term of three
numbers in harmonical series.
If a and b represent two numbers, and H their harmonical mean,
the corresponding arithmetical series is , ^, ^. The common dif
.11 !!,«,.
ference is ^ ~ o* ^^ 6 ~ H' ^*^^'^'®^»
D —  = T  ^. Hen'ce (solve for //),
25
386 ELEMENTS OF ALGEBRA.
To Find the Harmonical Mean Between two Numbers.
H  ^^^ (i)
Example L Find the harmonical mean between : (1) ^ and ^ ;
(2) X j y and x — y.
Process. (1) Here, a = \ and b — ^j^.
Substitute in (i), H = \
(2) Here, a = x \ y and b = x — y.
X^  7/2
Substitute in (i), H— '— .
Example 2. Insert three harmonical means between f and ^^.
Process. The terms of the corresponding arithmetical series are
I and J^.
Here, a = , I = ^^, and m = 3.
Substitute in (v) (Art. 156), d = ^.
The three arithmetical means are ^, ^, ^.
The required harmonical means are y\, f , ^.
Exercise 141.
1. Find the 8th term of IJ, l^f 2^2^, ....
2. Find the 21st term of 21 llf, 1^^, ....
3. The 39th term is y^, and the 54th term is ^. Find
the series.
4. The 2d term is 2, and the 31st term is ^*y. Find the
first six terms.
Insert :
5. One harmonical mean between 1 and 13.
6. 3 harmonical means between 2f and 12.
HARMONICAL PROGRESSION. 387
7. 4 harmoiiical iiu'iiii.s buLu uuii  iind ^.
8. 6 harinouical means between 3 and ^.
9. The arithmetical mean of two numbers is 9, and the
harmonical mean is 8. Find the numbers.
10. The difference of the arithmetical and harmonical
means between two numbers is 1. Find the numbers; one
being three times the other.
11. Find two numbers such that the sum of their arith
metical, geometrical, and harmonical means is 9, and the
product of these means is 27.
12. The arithmetical mean between two numbers ex
ceeds the geometrical by 2^, and the geometrical exceeds
the harmonical by 2. Find the numbers.
13. The sum of three terms of a harmonical series is 37,
and the sum of their squares is 469. Find the numbers.
14. The sum of three consecutive terms in harmonical
series is 1^, and the first term is J. Find the numbers.
15. Arrange the aritlimetical, geometrical, and harmoni
cal means between two numbers a and h in order of
magnitude.
16. If 50 potatoes are placed in a line 3 feet from each
other, and the first is 3 feet from a basket, how far will a
person travel, starting from the basket, to gather them up
singly, and return with each to the basket ?
17. There are four numbers in geometrical progression,
the first of which is less than the fourth by 21, and the
difference of the extremes divided by the difference of the
means is equal to 3 J. Find the numbers.
388 ELEMENTS OF ALGEBRA.
CHAPTEK XXVII.
RATIO AND PROPORTION.
162. The Ratio of two numbers is their relative magni
tude, and is expressed by the fraction of which the first is
the numerator and the second the denominator.
Thus, the ratio of 10 to 5 is expressed by the fraction ^^ ; the
ratio of I to I is expressed by the fraction f r f (= y%).
The ratio of two quantities of the same kind is equal to the ratio
of the two numbers by which they are expressed.
Thus, the ratio of $5 to $6 is  ; of 15 apples to 3 apples is i/ ; of
3f feet to 5^ feet is 3f ~ 5 = ^.
, The Sign of ratio is the colon :, ^, or the fractional
form of indicating division.
a
Thus, the ratio of a to 6 is expressed by a : b, or a f 6, or t, any
one of which may be read "a is to 6/' or "ratio of a to &."
The Terms of a ratio are the numbers compared. The
first term is called the antecedent, the second the conse
quent, and the two terms together are called a couplet.
A ratio is called a ratio of greater inequality, of less
inequality, or of equality, according as tlie antecedent is
greater than, less than, or equal to, the consequent.
An Inverse Ratio is one in which the terms are inter
changed ; as, the ratio of 7 : 8 is the inverse of the ratio
8:7.
A Compound Ratio is the product of two or more simple
ratios; as, the compound ratio 2 : 3, 5 : 4, 15 : 6, is 150 : 72.
RATIO AND PROPORTION. 389
NotM : 1. A quantity may be detiaed as a definite portion of any magni
tude. Thus, any definite number of dollars, poimds, bushels, acres, feet, yards,
or miles, is a quantity.
2. To compai*e two quantities they must be expressed iu terms of the same
unit. Thus, the ratio of 2 rods to 9 inches is expressed by the fraction,
16J X 2 X 12 396
163. Evidently the ratios 4 : 5, 8 : 10, ^ : Yi are equal to each
other. Ill general,
a ma „
I. If the terms of a ratio are multiplied or divided by
the sanu number, the value of the ratio is not changed.
The ratio 9 : 7 is compared with the ratio 4 : 3 by comparing ^
and . ^ = ^\, and  = ^. Therefore, 4 : 3 is greater than 9 • 7.
Hence,
II. Ratios are compared by comparing the fractions that
represent them.
If to each term of the ratio 5 : 4 we add 16, the new ratio, 21 : 20,
is less than the ratio 5 : 4, because \ is greater than f ^. If to each
tenn of the ratio 4 : 5 we add 16, the new ratio, 20 : 21, is greater
than the ratio 4 : 5. Hence,
III. A ratio of greater iiuquality is diminished, and a
ratio of less inequality is increased, by adding the same
number to both its terms.
If from each term of the ratio 32 : 30 we subtract 24, the new
ratio, 8 : 6, is greater than the ratio 32 : 30. If from each term of
the ratio 28 : 30 we subtract 16, the new ratio, 13 : 15, is less than
the ratio 28 : 30. Hence,
IV. A ratio of greater inequality is increased, and a
ratio of less inequality is diminished, by taking the same
number from both terms.
390 ELEMENTS OF ALGEBRA.
a c e g
Suppose ^ = ^=^=^ = r.
Simplify, br — a, dr = c, fr — e^ hr = g.
Add the corresponding members and factor the result,
{b\d+f+h)r = a + c + e\g.
Therefore, ^ = 6 + ^ t/t I = ^ = "^ =}" f' H"^'"»
V. In a series of equal ratios, the sum of the antece
dents divided hy the sum of the consequents is equal to any
antecedent divided by its consequent.
Notes : 1. The sign : , is an exact equivalent for the sign of division ; and is
a modification of r .
2. A Duplicate Batio is the ratio of the squares ; a Triplicate, of the cubes ;
a Subduplicate, of the square roots ; a Subtriplicate, of the cube roots of two
numbers. Thus, a^ : b^ ; a^ : b^ ; Va : Vb; fa : Vb are respectively fhe
duplicate, triplicate, subduplicate, and subtriplicate ratios of a to b.
Example 1. Find the ratio compounded of the duplicate ratio of
2 a a2 _
T : T2 V 6, and the ratio 3ax : 2by.
Process, ihe duplicate ratio oi 7 : Tg yB is r^ : r^ •
, . 4a2 6a< I2a^x ]2a*by
Ihe compound ratio vg : p ? ^ax : 2by,is — p— : 74 — •
I2a^x \2a^by ISa^x 12 a^hy bx
But — To — : TT— ^ = — To i u — = — = bx : ay.
b^ 6* b^ b^ ay ^
Example 2 If 15 (2 22 — y'^) — ^ xy, find the ratio x : y.
Process. From the given equation, a:^ — jV a: ?/ = ^ 2/^.
Complete the square and solve for x, a; = ^ ?/, or — f i/*
X
Therefore,  = a or  1 .
Exercise 142.
Find the ratio compounded of :
1. The ratio 2 a : 3 &, and the duplicate ratio oi^b'^'.a'h.
2. The subduplicate ratio of 64 : 9, and the ratio 27 : 56.
RATIO. 391
3. The duplicate ratio of 4 : 15, and the triplicate ratio
of 5 : 2.
4. 1 — 3^ : 1 + y, X — X i/ : I + s^, and 1 : x — x^.
a\h g'+fe^ {a^l^f a2_9^^2Q , a813a + 42
Simplify each of the ratios :
6. 5ax:4:x; li5xy:20a^; 2x^y:\x^.
7. iaxy.z^ay^; ^ ^ :a^na^.
Arrange the following ratios in order of magnitude :
8. 5 : 6, 7 : 8, 41 : 48, and 31 : 36.
9. a ^ b : a \ bf and a^ — I? : a^ + b^, when a > 6.
10. For what value of x will the ratio 15 + a::17 + ic
be equal to the ratio 1:2?
11. Find X \ yy if a:^ + 6 ?/2 = 5 a; y.
12. Find the ratio of x to y, if the ratio 4a; + 5y : 3a;— y
is equal to 2.
13. What number must be added to each term of the
ratio a : h, that it may become equal to the ratio m : nl
14. What number must be subtracted from the conse
quent of the ratio a : b, that it may become equal to the
ratio m : 7l1
15. A certain ratio will be equal to 2 : 3, if 2 be added
to each of its terms; and it will be equal to 1 : 2, if 1 be
subtracted from each of its terms. Find the ratio.
392 ELEMENTS OP ALGEBRA.
16. If a : 6 be in the duplicate ratio oi a \ x : h + x,
find X.
17. Show that a duplicate ratio is greater or less than
its simple ratio, according as it is a ratio of greater or less
inequality.
PROPORTION.
164. A Proportion is an equality of ratios. Four num
bers are in proportion, when the first divided by the second
is equal to the third divided by the fourth.
a c
Thus, if T = 7 , then a, h, c, d, are called proportionals, or are said
to be in proportion, and they may be written in either of the forms :
a : b :: c : d,
read, "a is to & as c is to d! ;"
or a : b = c : d,
read, "the rat'o of a to & is equal to the ratio of c to d;"
a c
^^ b=d^
read, "a divided by b equals c divided by d."
The Terms of a proportion are the four numbers com
pared. The first and third terms are called the antecedents,
the second and fourth terms, the consequents; the first and
fourth terms are called the extremes, the second and third
terms, the means.
Thus, in the above proportion, a and c are the antecedents, b and
d the consequents, a and d the extremes, b and c the means.
Note 1. The algebraic test of a proportion is that the two fractions which
represent the ratios shall he equal.
PROPORTION.
Let
a:b :: c :d.
By definition,
a c
~b^d'
Free from fractions,
ad = bc. Hence,
393
I. In any proportion the product of the extremes is equal
to the product of the means.
Note 2. If any three terms in a proportion are given, the fourth may be
found from the relation that the product of the extremes is equal to the
product of the means.
Let ad = be. \
a c
Divide by 6 rf, b~d'
By definition, a:b::c:d. Hence,
IL If tJu product of tivo numbers is equal to the pro
duct of two others, either two may he made tlie extremes
of a proportion and the other two the 7neans.
A Mean Proportional is a number used for both means
of a proportion ; as, h, in the proportion a :b ::h : c.
A Third Proportional is the fourth term of a proportion
in which the means are equal; as, c, in the proportion
a : h :: h : c
Ia'I a : b :: b : c.
Therefore!., b^ = ac.
Extract the square root, b = \/a c. Hence,
IIL A mean proportional between two numbers is eqv^
to the square root of their product.
Let a :b :: c : d.
Therefore 1., ail — he,
a b
Divide by c rf, ~ ~ d'
By definition, a. cy.b . d. Hence,
394 ELEMENTS OF ALGEBRA.
IV. If four numhers are in proportion, they will he in
proportion hy alternation; that is, the first will he to (he
third, as the second is to the fourth.
Let a :b :: c : d.
Then I., bc = ad.
Divide by a c, ::=_.
•^ a c
By definition, b : a :: d : c. Hence,
V. If four numhers are in proportion, they will he in
proportion hy inversion; that is, the second will he to the
first as the fourth is to the third.
Let a:b::c:d.
By definition, h~ d'
a c
Add 1 to each member, t+ 1 = ^ + 1,
a + b c^d
b ~ d
Therefore, a + b : b :: c + d : d.
Hence,
VI. If four numhers are in proportion, they will he in
proportion hy composition; that is, the sum of the first
two will he to the second as the sum of the last two is to
the fourth.
Let a : b :: c : d.
. a c
By definition, J ~ d'
Subtract 1 from each member,
a c
a~b c— d
«^ ~r = ~d"
Therefore, a — b:b::cd:d. Hence
PROPORTION. 395
VII. Jf four niimhers are in propoi'tion, they will he
in proportion hy division ; that is, the difference of the first
two will be to the second as the difference of the last two
is to the fourth.
Let a :l ::c '.d.
a 4h c + d
Then VI., = ^ •
also VII.,
c
a — b c — d
Divide,
b c
a+b c+d
a — b c — d
By definition, a + b : a — b :: c + d : c — d. Hence,
VIII. If four numbers are in proportion, they will he
in proportion hy composition ajid division ; tliat is, the sum
of the first two will he to their differeiue as the sum of
the last two is to their difference.
Let a :b'.:c :d,
e:f::g:h,
k : I :: m : n.
T»,/... ^ c e g k m
By definition, l = d'f=h'l=n'
Multiply the corresponding members of the equations together,
aek c gm
Ff'l'^dir^'
By definition, a e k : b f I :: c g m : d h n. Hence,
IX. The products of the corresponding terms of two or
more proportions are in proportion.
Let a:b :: c : d.
d c
By definition, h~ d'
Raise each member to the nth power, fcS ~ ^ *
396 ELEMENTS OP ALGEBRA.
Therelore, a" : 6" : : c" : d\
1 1
Extract the nth root of each member, — = ^ •
Therefore, a" : 6^ :: c" : cT. Hence,
X. /?i a^iy proportion like 'powers or like roots of the
terms are in proportion,
A Continued Proportion is a series of equal ratios ;
As, 8:4::12:6::10:5 ::16:8; a : b :: c : d :: e :f::g:h, read
"a is to 6 as c is to d a.s e is to/ as g is to A."
Kote 3. Four numbers are said to foiin a continued proportion when each
consequent is the antecedent of the next ratio ; as, a : b :: b : c :: c : d.
Let a :b :: c : d :: e :/ :: g :h.
a c e g
By definition, ^ = ^^ = ^=: ^^ •
a ^c ^ e + q a c e g
ByV., (Art. 163), l^dv f+h^ h^ d = r'h
Therefore, a\c\€ + g'.h + d +f hh i: a : b. Hence,
XI. In a continued proportion the sum of the ante
cedents is to the sum of the consequents as any antecedent
is to its consequent.
^^ a2+62 ab + bc
Example 1. If  .— r i> ^ = ~"a2~i — 2 » prove that 6 is a mean
ab + be 6^ 4 c^ ' '■
proportional between a and c.
Proof. Free the given equation from fractions, transpose and
factor, (b^ — acy — 0, or b^ = ac.
Therefore II., a : h :: b : c.
Example 2. If a . b :: c : d, prove that m a'^ +^p b'^+ nab : mc^
ipd^ + ncd :: b^ : d^.
iVI.,
a b
c~ d
a« ab
7^=7d'
a* b»
ab
nab
a^ n (t li
cd
ncd'
or
c2 ~ 'nc il
b^
pb^
a^ pb^
d^'
~ pd^'
or
c^~ pd^
PROPORTION. 397
a b ^1 V
Proof. From the giveu proportion VI., c~ 7l ^
a
Multiply by  ,
Square Loth members of (1),
By I. (Art. 163),
Also I. (Art. 163),
Also I. (Art. 163), r» = ^a
m a* pb^ _ n ab _ b'^ a'
Hence, ^^ =^ = ^^^ = ^ = ^.
By V. (Art. 163), ^...^'pd^^ncd = d^^'
Therefore XL, ma^ + pb^ ^na b :mc^ + pd^+ ncd :: b^ : d^.
Example 3. Find x when ^m\x\ ^mx . ^m^x^m—x
::n ; 1.
Procefw. By VIII., 2 ^m\x : 2 ^mx :: n+l : n1,
or I. (Art. 163), /^/^Ta: : ^m=^ :: n+l : n1,
By X., m + x : m x :: (n+l)» : (nl)«.
By I., {m + x)in 1)» = (mx)(nM)«.
Simplify, transpose, and factor, 2 n {n^ + 3) x = 2 m (3 nH 1).
_m(3n« + l)
Therefore, x  ^(^8^.3) •
Exercise 143.
J( ad = bc, prove that :
1. d :b :: c : a; d : c ::b : a; h : a :: d : c.
2. hidr.a'.e; c:a::d:h; c:d::a:h.
398 ELEMENTS OF ALGEBRA.
Find a mean proportional between :
3. 2 and 8; 3 and 1 J ; Handf; 8 and 18; a^bsmdah^.
4. (a + hf and (a  hf ; 360 a* and 250 d^ h\
Find a third proportional to :
5. I and f ; I and  ; .2 and .4 ; 2 and 3 ; f and f
6. 1 and VI; (a  hf and a^  62; ? + ?^ and  •
Find a fourth proportional to :
7. 2, 5, and 6 ; 4, , and  ; f , f , and f ; a, ak, and 6.
8. a3, ah, and 5a22,. ___, _______ and ^g^
If a : & :: c : 6?, prove that:
9. a + h : a :\ c \ d : c\ a — h \ a w c — d : c.
10. ac :b d :: c^ : d!^; ab : cd :: a^ : c^.
11. 2a + 3c:3a + 2c::26+3f^:3&+2c?.
12. 3 a  5 6 : 3 c — 5 c? :: 5 a + 3 6 : 5 c + 3 c^.
^oo A7 o ^7^^ + ^ r^, — 6 a b
13. fa:?.::.:H; ^ = __^ . _ = ^ .
U. 3 a + 2b : S a  2b :: 3 c { 2d : 3 c  2 d.
15. la + 77ib : pa + qb :: I c + md : p c + qd.'
16. ft3 . 2,3 .. ^3 . ^3. ^2 . ^2 .. ^2 _ ^2 . ^2 _ ^2
17. a2 + c2 : a& + C6^ :: a6 + C6? : b^ + ^2.
18. V^r:r^:V6::VM^: V^; l = \^P^'
PROPORTION. 399
If 6 is a mean proportioual between a and c, prove that;
If « : 6 :: c : rf :: 6 :/, prove that :
on J. . i 7.i^_L/ O'+'^c^Se 2a43c+4g
20. a:fe::a + . + .:^ + ^+/; ^^2rf^3/ = 26+3rf+4/ '
21. rf is a third proportional to a and 6, and c is a third
proportional to b and a, find a and 6 in terras of d and c.
li m + n : m — n :: X \ y : X — y, prove that :
22. ^ •\' w? \ ^ — iv? \\ y^ ■\ n^ \ y^ ^ n^.
Solve the following proportions :
23.
24.
a:3 ~ yS : (a;  y)8 :: 19 : 1 and a; : 6 :: 4 : y.
25.
If = = , show that a + b{c = 0.
X — y y — z z — X
26. A and B engage in biisine^Jj with different sums.
A gains SloOO, B loses $500, after which A's money is to
B's as 3 to 2 ; but had A lost $500 and B gained $1000,
then A's money would have been to B's as 5 to 9. Find
each man's investment.
27. Show that the geometrical mean is a mean propor
tional between the arithmetical and harmonical means
between the two numbers a and b,
28. When «, &, c, are in harmonical progression, show
that a:c::a — 5:& — c. Hence, of three consecutive
terms of a harmonical series, the first is to the third as tht
first minus the second is to the second minm the thirds
400 ELEMENTS OF ALGEBRA.
29. Find the ratio compounded of the ratio 3 « : 4 6,
and the subduplicate ratio of 16 &* : 9 a*
30. If  = 3i, find the value of ^ ~ '^ ^ 
y ' 2x by
31. If 6 : a : : 2 : 5, find the value of 2aZh:Zh—a.
32. If T = T, and  = ^, find the value of ^7 ^^ •
6 4 2/ ' 4:hy — lax
33. If 7 m — 4 71 : 3 wt + ii : : 5 : 13, find the ratio
m : n.
34 If s s— = TT , find the ratio w : n.
m2 + 7l2 41
35. If 2 a: : 3 y be in the duplicate ratio of 2 a? — m : 3 ?/
— m, find the value of m.
a c m
36. If  =  = , prove that each of these ratios is
equal to ^WHH^L
4 m^c
37. If 2a + 3Z» : 2,a  36 :: 2^2+ 37i2 : 2^2 3ii2,
show that a has to h the duplicate ratio that m has to 7i.
38. A railway passenger observes that a train passes
him, moving in the opposite direction, in 30 seconds ; but
moving in the same direction with him, it passes him in
90 seconds. Compare the rates of the two trains.
Solve the following proportions :
39. Vx + Vh : Vx  Vh '.: a :h) 2"^' : 22 ; : 8 : 1.
(x + y.x
: X — y '.'. m + n \ m — n*
40. ^ „ o 2 2 . 2 1
APPENDIX,
COMPUTATION OF LOGARITHMS.
Since the logarithms of all composite numbers are found by add
ing the logarithms of their factors (Art. 122), it is only necessary to
compute the logarithms of prime numbeis.
The following method for computing logarithms is the one that
was used when our tables were first made, although it is not the most
expeditious method now known.
Example 1. Find the logarithm of 5.
Since 10«> = 1,
and 101 = 10 (1)
and as 5 lies between 1 and 10, its logarithm must lie between and 1.
Extract the square root of (1), 106 = 3. 162277+ (2)
As 5 lies between 10 and 3.1622774 its logarithm lies between
1 and .5.
Multiply (2) and (1) together, I0i» = 31.62277 f.
Take the square root, 10 '» = 5.6234134 (3)
5 lies between 3.162277+ and 5 623413+ , and its logarithm must
lie between .5 and .75.
Multiply (2) and (3) together, lO^^ = 17.7827895914+.
Take the square root, 10«26 = 4.216964+ (4)
Since 5 lies between 5.623413+ and 4.2 16964+ , its logarithm
must lie between .75 and .625.
Multiply (3) and (4) together, take the square root of the result,
and we have 1()«876 _ 4.869674+. Continuing thi; process to 22
operations, we have, 10«8»7(h = 5.0000(K)+.
Therefore, log 5.000000+ = .698970+.
402 ELEMENTS OF ALGEBRA.
Example 2. Find the logarithm of 2.
log 2 = log i^ = log 10  log 5 = 1  .698970 = .301030.
Examples. Find the logarithm of 11.
101 = 10 (1)
• 108 = 1000 (2)
Extract the square root of (2), lO^^ = 31.62277+ (3)
Multiply (3) and (1) together, lO^s = 316.2277+.
Take the square root, lO^^s = 17.78278+ (4)
Multiply (4) and (1) together, 10226  177.8278+.
Take the square root, 10ii25 = 13.33521+ (5)
Multiply (5) and (1) together, IO2125 _ 133.352I+.
Take the square root, 10i0625 = 11.54782 (6)
Multiply (6) and (1) together, 1020625 _ 115.4782+.
Take the square root, 10i03i25 _ io.74607+ (7)
Multiply (7) and (6) together, 10209375 ^ 1 24.09368+.
Take the square root, ioi.o46875 ^ 1 1.13973+ (8)
Multiply (8) and (7) together, 102078125 ^ 1 19.70845+.
Take the square root, 10io390626 _ 10.94113+.
Therefore, log 10.94113+ = 1.0390625.
Continuing the process, the logarithm of 11 maybe found with
sufficient accuracy.
Example 4. Find the logarithm of 3.
Take 10^ = 1 and lO^ = 3.162277+, and proceed as before to 14
operations, and we have log 3.0000+ = .47712+.
A table of logarithms to four decimal places will serve for many
practical purposes. In the tables most generally used by computers
they are given to six places of decimals. Seven to ten place loga
rithms are necessary for more accurate astronomical and mathemati
cal calculations.
ANSWERS
TO THE
ELEMENTS OF ALGEBRA
BY
GEORGE LILLEY, Ph.D., LL.D.
EXPRESIDENT SOUTH DAKOTA AGRICULTURAL COLLEGE
'TEACHERS' EDITION
SILVER, BURDETT & COMPANY
New Yobk . . . BOSTON . . . Chicago
1894
Copyright, 1893,
By Silver, Burdett and Company.
John Wilson and Son, Cambridge, U. S. A.
ANSWERS
TO THE
ELEMENTS OF ALGEBRA
Exercise 1.
1. a plus 100 ; a plus 10, minus 2 ; etc.
4. q plus t, plus 8 multiplied by m ; etc.
6. m\n\r~t] etc.
7. m ^ n \ b\ m + n — b.
10. X — m ^ n marbles.
11. 7i + a; f y 4 6 4 w.
12. arrti\n — k — X — 1/.
Exercise 2.
2. A; times ^, plus m times A; divided by c times w, plus a
divided by 6; or, to the product of k and /, add the quotient
obtained by dividing the product of m and k by the product
of c and n, and to this result, add the quotient obtained by
dividing ahy b; etc.
Exercise 3.
1. xyz; 5mn; Sxyj 15abmn.
^ ab ab 25 mn r ,
2. — jT, T. 3. ■ , etc.
a \ b a ^ b m { n
ANSWERS TO THE
a mn
4. 10. ^. ^. 20 amn.
o o
6. 18 mi'. 18 m 71 + 18 m r.
mn
8. aoc \ mnr, 13. 9. .
G
11. 25. ^. 14. 5. ^»' + ^^'
m r
dt dt dt
15. It. \ bt. r .
n n n
Exercise 4.
1. m fifth power; 3, m fifth power, x second power;
etc. ; a, h second power, plus h ; etc.
3. 10, ah tenth power; m third power, n third power,
m n third power, etc. ; the third power of m second power
minus 3 n.
4. Etc.; 3, a second power, b, times the third power
of a minus b second power ; a second power plus b second
power, times the second power of a third power minus b
third power.
6. 10 m, plus n fourth power, times the fourth power
of 10, n second power, minus m fifth power is less than
15 a times the second power of x, minus y second power,
times the third power of x plus y ; etc.
8. m + n. 2x. (a + by. (x — y)\ 5 (x — yf.
9. (x' + yy. (icH 7/2)2. m^x^y^. ^xy^. {x'' + y"") {x^  y%
11. 1 m'^n^—2n'm^+^a^h'^+^a''h^ + 5a\ '/a\b^
m — n, .'. (a + by = (m — ny.
12. .'.X = m^,'.'x + 3m^ = 2x + 2m\ a + a + a
to ?i — 2 terms = (n — 2) a; etc.
ELEMENTS OP ALGEBRA.
Exercise 5.
1. 3; 448; 60; 180; 64; 9375; 390,625; 1792; etc.
2. 144; 60; 64; 250; 4; 40; etc.
3. T^; 3; 7; 75; 32,400; 288; etc.
4. 6; 60; 2; 3; 5^; 72; 0.
5. 9; 160; 2048; 81; 8 ; 13^; ^; H
8. 20; 11. 12. 2; 0. 16. 4§f ; 2^.
9. 0; 12. 13. 249; 134. 17. 1; .
10. 14; 6. 14. 22; ^. 18. 5; 58^.
11. 18; 14; 2. 15. ^; 36. 19. i; I4.
Exercise 6.
1. 11, negative. 11, positive, etc.
2. is 6 units greater than — 6 ; etc.
3. 6, 3, 9, 12, 11. 10. ba.
8. 2 times the expression in brackets, 3 i in the posi
tive series plus 5 a in the negative series, and from this
result subtract 6 times the expression in brackets, a in the
negative series plus b in the positive series ; etc.
10. Value, 26.
11. Value, 1 + ( a:^^) + (+ a;«) + (x),
12. Value, 2. 19. 80; 624^^. 26. 56; 15.
13. Value, 0. 20. 3; 15^^; 1^. 27. 2; 15.
14. Etc. 14§^§. 21. 9; 2; 15. 28. 102.
15. Etc. 17. 22. 127; 21; 6; 1. 29. 52; 18^^.
16. Etc. 6. 23. 3 ; 6. 30. — 15 ; 12.
17. Etc. ^. 24. — 4 ; 6. 31. 2 a + b.
18. Etc. 2r»^. 25. 16; 55.
ANSWERS TO THE
32. 5« + (6  1). 33. h + ^±_^.
34. x\x{x{....toa terms, or ax.
35. n, n { 1, n \ 2.
36. m, m — 1, m — 2^ w — 3, m — 4.
37. (a — 1) m ; (m + w) m.
38. X — S, X — 2, X — if X, X \ 1, X + 2, X \ 3.
39. a 2/". ^ , a;
40. 2x + '^ + abc, 41. ^.
b y — c
42. a" +  — 5 ( ~\ >h — x.
43. (cc^ — ^»« 4 2/'") ^ ^" < g'^".
a** « — ^>
44. x"" —
b"" a' + b^
(A
45. — ^ — .X ?/ + cc + X + cc + • . . • to w terms — a"*.
46. x"* H ■=x — y\ (4a+^» — m) + — .
47. 5^8 — 3 6i^Z>8 + 2^»2. 49. %x^y'^ — a^b\
48. 3^2 2cc«?/ + ^3. 50. + 6a(ir 4 2/^).
a: — y
Exercise 7.
1. (+15.1 a). 5. (5£c2). 9. (+3.8a^c(/).
2. (+15 ax). 6. (+/2^). ' 10. 7.71 (^' + c).
3. (+41c). 7. (21a8). 11. + 5.81 (x  ?/)8.
4. (+ 2 a ^ c). 8. ( 3.5 a" U^). 12.  3/^ () .
ELEMENTS OF ALGEBRA.
Exercise 8.
1. (6.69 a;) + (+3.5^). 4. (+2^ x') \ {+ i ab).
2. (+Ha) + (5'5«^) 5. 14.9 a 2.67 (xy).
3.
(+8a'a:) + (+7c«x'). ^ 6.1Q2.3Q.
Exercise 9.
1. 5 y. 2. 4 // — 5 a. 3. x. ^. a + b \ 2 x \ t/.
5. —33 ax — 4:bd \ 5ni7i.
6. 2a + ^ + c + </ + ^'i + a^.
7. — 8 a i c j 53 a 6 m — 20 c m. 8. 4 w.
9. 3 a h 3 /> + 3 c + 3 rf.
10. 4.25 a^ 4 3.3 f + 2 6. 11. 2{^^a+l^^ab.
12. iVo ^'^■^ + '^l ''^ — 2t\ w n — 2^.
13. — ac+1.92frf. 14. — (1*^.5 a^b+ .25 a b^{b\
15. 7.8 (m — ny — 6.03 (« + y)^ 16. a^ 6» + x^y,
17. 7tJ5 a^ + 5i « a^« + T^^ //« + 3^ x^ U
18. I «• 4 J\ Z^''  1.3 c»  «2^ + m a^c.2ac^^l}fab^
+ ljgga6c + 2*3c1.3^c'».
19. 21.33 a^ + 8.37 ar^  5 x f 8.5.
20. 2.6 a* b^ r» + 2.8 «« 6» c^ + 3.91 a^ 6» c*.
21. 6.09 a* + h f' + 4.97g 6 c + 3.5 c"* — 3.03.
22. \aX"l.Sab + 5.2x + ^xy.
23. 12.91 a + 4.1 y — 6.82 z.
24. 3 o 6 + 10 J y 4 9 a // — .T« y + 22.
25.  J(m3a)".
26. 5 a*. 29. <?» + ^.« H c« — 3 a 6 c.
27. 10a« + 86«+12a + 12. 30. 0.
28. x* — y*. 31. 4 a:'" + 2 a" — a".
ANSWERS TO THE
Exercise 10.
1. lOaHc; SSab^xi/, 7. ^ (x + y).
2. 2x'y^; 0. a 3ax\
3. —daxy, xy^ + bc. 9. 26axy\
4. —3.75x\ 10. %x^y.
5. 2.72 a bc^ 11. 8.9i(a + ^)2.
6. —13.1 VI np^^ 12. 4TV5ra^".
Exercise 11.
1. —2x + 4:y — 3z. 5. ~2x + y — :^z.
2. 2xy4.z. 6. ^.T + 2/ + i
3. —2a — 10c. 7. 2a^ + 3aH — 3hc—10,
^. Ix+^y + l^z. 8. 2x^y + 2abx + l.
9. abcxy — 3aby}5bx — 3.
10. 2^372/ — 5ac?/ — 2a^c + a + l.
11. .6 ic*  1.8 x^ + 7.03 cc  9.
12. —x^ — 1.2x* — 2x^ — 2x\ 2.
13. ^^ 77Z8 — I y«, 7^2 _ J s ^2 24. 7^8 — «.
15. ^^m^J^y + 2lin^x.
16. 2.5«2^,c — l^a?/" 4t^c. 17. 0.
18. 2 a;*" + 4 ;?;" y"* + y"^.
19. 1.8 ^/"^ x^  1.7 a ^»3 .X + .37 b"" c x"" + .7.
20. a^ _,_ 3 53 _ g ^4^ 21. 3.55 ic^ + 33. 2y^x + S.5x z\
22. ic'» + 2y"'. 23.  iy _ 1.9§ 2/^ + 17 ?/« + 3.
3. 22a^2a2.
4. 2 a ^ — ^2 _ ^2^
^2 n yin rjf.Zn 2 ^*"
Exercise
12.
1.
— 4 m^ — 2 w 
 5 7z2 + 6.
2.
4.x^Uy.
5.
— 2 m.
6. 4?/
wl
ELEMENTS OF ALGEBRA. 9
7. xhyiei/^ + Sz' ^1.
9.  ^ ari + 2.9 a;* yi  33 yl
10. 3.9 cyi + 1.6i ax + .31 ^»  1.2 7m  5^%.
11. 5.5 6" — 3.25 «"• — 4 w. 12. y — x^ — 1.2 i".
13. ^ c' — l.Ta'" — o^b"* — 2 d".
14. c»  i (a»  by  12 (x» + y2^«.
15. 2 a^ 21. 6 rt — 2 a ^.
16. 3 a;«* + 2 x^* — 6 a;". 22. aJ ^i.
17. 2^2;— I a^ 2. 23. 4y a^ ^a» — 4.5a + 3f.
18. \ //i« +1. 2^. llbc \l mn — ^xy.
19. 7 J/'' + 18y — 4. 25. 4 x*^ — 2x'''^.
20. 0. 26. x"— 4.
27. 64 (x 4 y)* + .3 a + x"*  3^ ax"  C + 3.
Exercise 13.
1. 7x8; 15a«x«; 2a'*^*x«.
2. ix"y»«m««"; 2an»c«<;»w»w».
3. a6»c»y"^*; ?a"x»y"«.
4. 30«"x*y"; ix'" + *y"+^
5. 2a^'/>^V^*»x^<>y"; Ti5a'Z»"r"j'" + »y"+».
6. vSa"+>6'' + ''x^+V"^*; ^'/A
7. a'" + 6»+»'; 5.7 x*y". 8. a»6»x«y«; a^ft^".
10. 2a»x»y". 15. .765a«mM+'x— "'1.
11. 3a*'+''6" ♦c'+'"rf*. 16. la'"+*'**^"//"+*«'".
12. 2a* + »c*++*rf«x'"+". 17. l\(a\hy\
13. 2 a*l m^ n** x*» /. 18. 3 (c + ^)><> (a + ^)".
14. 5 aHi a;*A yi. 19. ^ («  ^,)+ *» (x — y)'"+«.
10 ANSWERS TO THE
Exercise 14.
1. —2S10anG', a^x^'f. 6. 3aH^c^^d.
2. w'hx^y'; —ha^'^hU^d}. 7. an^(Px\
3. 4:aHcxy) .Sacm^x^T/^ 8. — .3 a^^ x^ yi ^ aH x^.
4. ^a^bcdx^yz; a^^+H'+^x^"" y\
5. — 3 a^b^x^yzvw, a'«+" + i^>" + 'a:"* + ^ ?/"+\
9. — m^ n x^t yH '^ — 2.1 ai bl 11. A X 3" mx^ y'^^'p q"".
10. a^'^ ''■ a aa to 10 ni3iGtovs. 12. _ 2^« + »"3^°« + ^" + i.
Exercise 15.
1. abU^+aHc^aH'c, ^ aH'*  :^ a^^^ c^  ^ aH^^ e^
2. 5 a^ Z*» c^o — a^ b'' c"" — 2a^b^ &'' ; .12 ic^ y^  .1 £c^ /
3. 3m^ n — 1\ w? ii^ + 3 m ?i^ ; 9? y — xy'^ —l\x^ y^.
5. 9 a«Z»ar3— «H^aji3^ .3a^>«ar«; p^ x'^'^r — p qrx'^^''
— pr^ ajtn
6. 3«'"^>2_2«^« + 4 rt»» + l^n + 2. .Ga^'^Z;^^  2 6/"'^^>2p
8. a^i ^^s  a^^ ^»^^  a^ b^^ + a^b; x^ yh  x^ y\ \ x^ y
— .Bx^yl
9.2 .T.3 ?/^ + 8 a^2y6 _ 8 _^ yv . ^„2 ,^o _ 2 ^^p ^2? ^ ^o ^^2,^
10. ^V «^ ^'^ ^^ + ^7 «* ^ cc' + ^ a^ Z*^ ic^
11. — cc§ ?/ + .^^ ?/^ ; ai ^'«^ — 6?i b'^\
12. 45a!i3i7/°i45£c'«^/3>.
13. ^V ^'^^'' — :\ ^'''^"■•'^'^' — 2 ^>'3\tV 4 ^ //75.
14. 180 aSm/.am _ 189 ^4m ^im _ 4Q ^2m Jl2m^
ELEMENTS OF ALGEBRA. 11
Exercise 16.
1. a*{aH^+b*; a* + 4 a^x^ + 16 x*.
2. x^ \ i/; ^xij.
3. 2/' + 'y'~lly'12y + 27; y* + a y» + « V
— a t/ z — a*z — a*.
4. \x*l,^^x' + ^jf', .64 a» 4 3.24 a ^2  2.7 ^»».
5. a^{2x'2x'i/ + x^xi/ 7/ {•!/',
l^x* 1^ ax« + i a^x^ — I a\
6. a;« + 8a:y + y — 1; H a:^ + 6 aic* + f a*x« + faa^*
+ 3a^x^ { la^x _ ^ a«x« — ^ a«a  j^jj a».
7. a'  7 a;« + 21 a;*  17 a*  25 a« 4 6 ^2  2 a;  4.
8. a» + 2^i«45a* + 2a24. 1.
9. x» h 1.25a;' + .25a;« + Sx*^ + .5.r* + .25 x' + 1.25 a:^ + 1.
10. 4 + 32 a  4 a^ f 25 a»  6 a* 4 «».
11. x» — 32 y*. 13. a« — 3 a ^> c + // 4 c».
12. x^* 4 y^^ \ x^ \ i/. 14. «» 4 3 rt ^» c 4 ^' — c*.
15. a*^»2  rt^2 4 ^*<^ 4 ^f/^. 16. — 8 a2 //.
17. .3 m« 4 2.90 « w — 3.01 i m  .3 71^ 4 3.01 a 7i  2.99 b n
— .01 a^ 4 .1 ^.
18. a^x^'* + b^x'^'* + 1^ \ 2a^»a;'" + " + 2ara:" + 26rx";
ari — yl
19. a*" + 2a'" 6 4 b^"', a*"* — i*; a:l 4 &^*+ J5a*4 ^'.
20. 6a;'" — 4a;;/'*~9ar— V* + 6y; a^am+a _^ ^^^,3.n+2
+ a^6x* — a^ar'"+' — i'^x^'^ff^^'^a:* — ax"" — ^x"
— a Ax.
21. 3a*'"'x43a2 + y + f(2''^'" — Sri^'^+^a; — 3a'*''?/ — a'"
43a«x2 43a«xy 4 «*"^; x5 — x=//? — xiy + y*.
22. .04a — .09 61; x — y. 24. l + x + x^ + x^ + x^ — x^
23. X* — i/\ 25. X* — 5 a* x* 4 4 a*.
12 ANSWERS TO THE
26. 120 ic*  346 x^  205 x^ + 146 a;  120.
27. x^ + x^ + 1. 28. x^ + a*x^+a\ 30. a^"*— Z.6«,
29. 3a^oa'bSaH''j7 aH^ + 6aH^2ah^ h\
Exercise 17.
1. a2^2a15; &2_^^>30; a;2^7a; + 12; a;23a;4
a;2  5 £c  14.
2. ^2— 14ic + 48; a2+4a45; a2_4^_32. 4a;218a;+20
9ic''^+ 6aj — 35.
3. :c6 — 7ic^2/^ + 12?/^; a;^ + xy — ^y'^; a^"^ + a"" — 2
9:i;i«27a;5 + 20.
4. 4a*/8ay32; 9 a^ a;2_^ 9 a ic  28 ; a;«  a a^^12a2
• x'o _ ^2 ^5 _ 6 ^2
5. 4a;2_2(ia;— 2^2; 4x2" + 4 a x*^ — 15 a^; 9^:2 — 60:3/
— 2 y^; Ax^ — Amx^ — 24: rn}.
6. a;2_6^^__5^2. ^2^3 ^ ^_4q^,2. ei^ _ 8 a^ £c + 12 a;^ 5
25x2»5a2;K^o_i2a^
7. 25 a:^«  25 cc5^2 ^ 6 2/^ 9 a^« + 3 a^ (2 a ft  4 a ^/)
%a'b'; «2" + a" (3  6)  3 Z».
8. 16^2 + 4a(^> c) ^»c; 2na' — 10a(b^c) + 4.hc',
a^y'^ + I aa;^/ + ii^^', «^ + i «' — I
9. 4ar + 26 a;^ + 12; 4 a + 2 a^ (5 — 3 a a;) — 3a ^»a;;
.09 a; 2/2"  .3 a;^ ?/" (a; + 2/) + «^ V
Exercise 18.
1. 4a;^92/^ a:^  4 2/^ 25 9x^ 25 a:'^  121.
2. 4 a;2  1 ; 4 a;2  25 ; 25 x^y'  9 ; c^  a^.
3. c* — a^\ m^ri^ — 1 ; a^ y^ — }p\ a^ x^ — 1.
4. x^ — y^; 1—p^q^; m^ — n^\ a^"* — a^".
5. 25x2?/216 2/^ 25a!^9y^ x^9x\
ELEMENTS OF ALGEBRA. 13
6. ^a'^x' — b'^y^', m^^ — ii^^, 100 a" 2"' _ 169 6"''*.
7. m + w; 16a — 400a;*; al — b%,
8. 121 a: 900 y; 225 a* *• 256 a 61.
9. \aH^tb^x^; a^b\
10. aH^l\ 16 a*'" — 256 a*".
11. 625 a»2  1296 b^ ; a" ^^ _ «« 6".
12. x\x'y^; ec'"^af^"*
Exercise 19.
1. 4 a** — a" + 6 a" — 9 ; 20 a^ — 3 x2« — a;' — 6 x^**.
2. 4 a*' 4 li a" + 3§ a^^ + 6 a'  15 ; 2 « + ^5 a;5   a;4
3. _ § a» + 6^ at — § a2 — 4^ al + 6 a* — a« + I al  a2 ;
6 a*"—'*' + 3 ar^b^f — 2 a^sp^sp _ ^p ^p
4. 25 a^V* + 20 a;2«2/26 _ 15 a:« y* _ 12 x« y% a^* + a"
+ a— 4 a*".
5. .09a«.156a*6 + .22a*6«.488a»6«1.39a«6*+.3a6*
+ .rib\
6. 1 _2a;J3a;i + 2aji + 2a;i; a\ — al ^ 4al8iai
+ ai  2 ai + i a»  i a'^ + ««  2 a"*.
7. 2a;t + 2x53x5a;l + a;84 3a;«; al"+ai".
8. x""+' — 2a:*"+'*' — 2ar2''+2 4.2a^'' + ' + a:''^* + 2 3f + '— a^+*
 2x* + x"'; j;2n+2_4^2n _^ 12a:2i — 9x''**.
9. 2a:»"*4a;"'' + 2.1ar«».9ar*'''+.lx*"+2+.2a;'»+*
4. 2a;»"» — Sar*— 2 + x«— » _ ar^.
10. 9af+"' — 34a^+— » 4 29 ar +  + » — «•"+"+*.
11. 2a;^+> — 63:*+* 4 2x«+'^ — 4a:'+* + 3a:*«+»  9 a:»«+*
+ 3a:*^+» — ^Q^'^''^ — 4a:»'^+» + 12x«'+*  4 a;'«+»
+ 8a^'+^
12. ITar^^V*"^*— lOx^^+V + l^x^'^V"^'— ^^"""V*"^*
14 ANSWERS TO THE
13. m^ + '— 3m^+'i 7i+ m^+'^Ti^— m'+'^Ti^ _ 3^p+r+i ^
+ 9 mP + ' n^ — 3 mP+''^ n^ + 3mP + '■ '^71^ { 7iiP^'+'' 71^
15. 2 x^y — 18 x^^^ + 6 a;3y _ 18 a:'^ / + 4 a:?/^
16. f — x^"*; ^^x'y'^b _^^^'zay2i,^
17. x^"— 2/2m. ^_^^_20; 49x2 — 9^2,
18. 16 x2 — 8 x0  15 a:2 ; ^cH^ — ^\ ^» Z/"* ; a^n _^ ^14 ^n
+ 499a2«.
Exercise 20.
1. 16aH^', 21a>^m', 32x^y^', .00032 a^o^^i^c'O; .01a2»^,2«.
2. 49a«^^ 121 a2&4c«<^^;  27 c^ a:« 5/I2 ^i^ . 27a^^^»«y^
25 a^^^^^^yo ^20^
3. 256 a« 6« ci« ^^^4 ^^32. ^10 ^10 ^10 ^10 ^10. _ ^^9 ^e ^s . ^8 js.
729a^6^8c^; ^aH\
^4nj.3n^«^2n^
5. — 8a^'7»'«"; m"=; ««'; a^b'^c^^', m""*'/!"*'"; 8; a^n^ 1.
6. 16a*5«c*?^^; ri^^ + i^smn^ ^»»« + iam. ^rri'n^x^^f.
7. 54am3 7i»a:^2. 81 a^« 5«^ c^* m^^ ; 81 a^^,*. a«^« + i.
243a^20c'#ic*?/§.
9. _ ^14^21 ^56^28^. _a^rm,/»5 ^'"'b^'k^*; _8a2i^,i4. x^y^\
10. X^2«^42m. ^21^65. 4" ^7n J7n ^7»« . _^7m»n«. J^Uk ^Ukm^
11. m" (a — 3 <^)P" (x — 7/)'" ; (a — 3 c^)i"' (a; — 2/)^" ;
3" (a — & + c + c?)" (a — cc)""*.
12. a" ^>" c" (a — &)"*"(cc + y + z'^Y" ; ct"' (x — y + ^)^"' (cc — y"»)8'»*.
ELEMENTS OF ALGEBRA. 16
Exercise 21.
1. a;« + 4 X + 4 ; m^ \ 10 m + 25 ; n^ + 14 w + 49 ;
a«20tt + 100; 4a;* + 12a;y + 9y^ a"" \ 6ab + 9b^;
2. J2+ 10xy425y2; 9x«30a:?/ + 25 2/'^; 4^2 + 4^26
+ a62; 25x230x^^ + 9x2^2; 25a2^,2c'2— lOaic^ + c*;
a;iyi_4icy + 4?/*; a^m ^ Qa'^b" { ^b'^\
3. 4 x^ 4 12 a* X + 9 a ; x^ f/^ + 2 x^ y { xt , 9 a*
+ 30a»6» + 25ai»^»«; l2x + x2. l_2cy+cV;
m22w + l; a2^,4_2a^^+l;  a^ /^ a* + ^i^.
4. ^a2^*f 3a6»x» + 62x2; j^*'*^*' — 2j9*^'"r«4 r^';
.00000004x2'" + .000002 x*"//" + .000025 y2».
5. 3»j/wMtV — 2w»7i'' + 'y/ + Y^^''^^"'; icV + 2y2a;«
+ //2^2_^3.ag2 _j_ 2ar2y^ I 2xy«2. 4x* + 5x«+l
+ 12x»6x; x* + 6x2+ l_4x» — 4x; x*4x2
+ 16 + 4 x»  16 X.
6. 4a?*H13x« + 94x»6x; x« + 25x* + 4  lOx*
— 4 X* — 20 X ; 16 w* 4 w* ^2 + 7i» 4 8 m^ n* — Sn*
— 2m^n^\ x» + 9x2 + 4__6x« + 4x* — 12x.
7. x2 y2 __ 4 ^8 ^ 1 _ 4 ,j jp y _l_ 2 a; y — 4 ?i ; rti^ ■\ v} ■\ j^
{ q^ — 2 m7i — 2mp — 2mq\ 2np \ 2nq \ 2pq]
a;« + 8x^ + 16x2+ 94x'^14x^12x; 1 + 3x2
+ 3x* + x« + 2x + 4x« + 2x»; x2 + 9^/2+ 4^2 + ^2
+ 6xy + 4ax — 2/>x + 12 ay — 6 bi/ — 4ab.
8. 4x*+5x* 17x2 + 9 + 18x»6x; x2+4y2^9^2
+ 4n2 — 4xy— 6x2 + ^ n x \ 12 yz \ ^ n y — 12 nz\
^2" + n2"* + />2» ^ ,^2« ^ 2 ?;^'•7^"• + 2??^''/>" — 2m* q"^
+ 2n'/)» — 2n^'" — 2/>»^'"; ^a^ + g^ + f — 2a6
— a + 9 ft.
9. 4a« + 4 62+TVc2~2ai + iacAc; x2" + y2m _,. ^ ^2
+ i 62 _ 2 X* 2r + <» X  2 6x* — ay» + ^ *2r — J\ aft;
^.r* + 3x2+ $x«"3x; a:«+}x«+^2x»
+ ^x*§x».
16 ANSWERS TO THE
10. 1f ia:+^V^'; i^'^^' + ^Vi^' + Aa^; ^^a'^
4 a;*3 + 25 x + 49 + 20 a;i + 28 ici + 70 o^i
11. 9x + 4.x^ + ^xi{xi~12x^ + 2a;i — 6a:J — £ci5 44ccs
523 x3i
Exercise 22.
1. a'7 a«^» + 21 a^^2 _ 35 ^4^8 ^ 35^8^4 _ 21 ^2 ^5 ^ 7^^.!
— ^>'; a^ + 6 a° ic + 15 a^x^ + 20 a'^c^ + 15 a'ic* + 6 aa:^
+ a:«; a^4:a'c+ 6 a^ c''  4: a^ c^ + aU^; a^16a«
+ 96a«  256a» + 256; 16 + 32a + 24.a'' + Sa^ + a^
a^  5 a* + 10 a^ _ 10 ^2 _^ 5 a  1 ; 1  5 a + 10 a^
~10a» + 5a^a^ ; 16 a* 96 a^i + 216 a^b^  216 ab^
+ Slb\
2. ic212ici+ 54ic108xi+81; a^a:^ 15a4a;« + 90 a^a;'
 270 a'x^ + 405 ax^  243a;^«; x'  15 x' + 90ic«
270a;2 + 405ic243; Sa^x^ma''Px''i/^54:aH'xy''
 27 ^« 2/«; 16 a* o:^ + 96 a^ ^ a;^ y + 216 a^ V x" y''
4 216ab^xy^ ■\ SI b'y\
3. a« 4 12 a^ + 60 a" + 160 r^^ + 240 a^ + 192 a + 64;
a«  12 a^ + 60 a' — 160 a^ + 240 w" — 192 a + 64;
16 — 3^ a 4_ I ^2 _ ^a^ ^8 _^ ^ ^4 . ^1^ ^4 _ 3 ^3 ^
+ V <*''*'  54.ab^ + 81 ^4 . _i^ «4 _^ ^ ^3 ^ _j. ^ ^2 ^2
 3^ a 6» + ^V ^^ ai« + 10 «» ^ + 45 V ^2 + 120 aH^
+ 210 a^ b^ 4 252 a^ b^ + 210 a* 6« + 120 a^ b' + 45 aH^
+ 10aP + b^\
4. a + 4 4 ai — 4a^2a«+4ai; 4x^+4^48:^2^2.
l + 3a2_^^4_,_2a4.2a^3; _l_5a24a* + 2a
+ 4a».
5. 16a8i + 64«63; _7 + 20a16ct2^4a«; 1 + 2 • 3^ • 5J
^ 2 . 2i • 3^,
ELEMENTS OF ALGEBRA, 17
Exercise 23.
1. 3a^bj 6ab\ ^abc^\ f mA.
2. w«; a^ aH^c^''', a?"; 2^'.
3. 5al AJa;^ ^ai**; 3a••^mx'^
5. (5c — y)2; (ac)*;  6*i r"' A;««.
6. Oa^^c^ a'"'"'; 2'm'w"^
Exercise 24.
1. 2a;; 2a*6*c«; oa*; 1.
2. —3a; ^a^i^c*; .Saft^c"^; SOa'^ir*.
» a A 5 m~ * a;* V* n i ^ 2 s
*• — ^^ «7wy«; — 31 ma: 2/^"*"'.
5. 6a:"'/'';  ii(a^)c"^ ^ai^^i*
6. 10 a;l y i (a;i —  y') A ;  w w x A y.
7.  4 a* (x  y)^ 2 1 ; m" n"" (a;  y)^'' (y  «)H.
8. 6a^6c; 2a'» + "6 + "c'.
9. 2aSil.
k 13. ^a^^d^^c*"".
11. 2a6*c*rf°x». 14. Ha"'6''icIx».
Exercise 25.
1. 1 + 3 ay — 4 aV; 3 m^ /i^  m 7i  2 + ? .
mn
2. 1 J«c aft r2 + aH*c^ 6a;*— V a: 4 40.
3. 4a«+ f a3 + ; 4a66^a^
a
4. 9aftr« — 12aft' — 5c*+ ^aftc.
2
18 ANSWERS TO THE
5. 3mn — rn^n"^ — 3 n { 5 th n^ , x"^ — x^^ ?/'. 
6. 2 a — 3 ^> + 4 c; — J3O ^^'^ + 2 n".
7. _ 3 ^4^7 _ I ^2^1 ^ 2 ai; .9 7i^e^  1.2 ?iA.
8. 36f 05 2/" + 10^8^; 3mi« + 160m80 7y^l7^.
9. ^m2 _ 2 a«>2 4. 3 a!'^\ m"2 — m**"^ + ??*" — w" ' *
10. — 3 xy""'' — 2d^x^ + 4.a'^x if''.
11. _ a»»i y^ 4. a"* ^> — a'*^; — f a^ ic'^ + \l a x^.
12. 6 a — ^ ^ — 6'; xi — 3 ars".
13. 3 m5 — f ?j3 + ^ r^if ?ii
14. 4 (x  ^)*  3 (x  y)2 + 2.
15. 2«'i + Sx"'?/" — 18a:'"y'*^'+\
16. (cc — ?/)'''* — m«".
17. (x + yf' (x — yy" + (X + yy'(x — 2/)«^
18. 3 7?i — 2 71 — 4 ; «i — a*^ ^>2 + ai
Exercise 26.
1. Tx'' + 5xy + 2y^
2. x^ — x"^ ?/ + £C z/'^ ; a^ + or ^ — b'^.
3. vy« _ 2 2/2 + 7/ + 1; 7/5 + 2/* + 2/' + 2/' + .V + 1
4. JC+22/.^; a7 + a6^_^^5^2__^4^34^3^44^2^5_^^56 4.J7^
5. 2 a; + 3 ^ ; a^
6. .25 x2 — 3 X ?/ + 9 7/2.
7. 27x2 + 12x7/+ 6 7/21; x^ — x^ 7/ + xif — y^.
8. 1 — 7/ — X + XT/; 2 7/3 — 3 7/2 + 2 7/.
9. x2 — X 7/ + X ;2 + 7/2 + 7/ ;>; + ^2 . ^.3 __ ^^ yj 4. /jjj 2/i + 7/i
10. X 7/ + 7/ ^ — X ^ ; X'^" + X^ 2/^ + aJ5 7/t + X? 7/5 + 7/t.
11. 4x2 — 6X7/ — 8 7/2. , ■ •
12. X2 — X7/ + X + 7/2 + 7/+ 1; ^X2^X + yV
ELEMENTS OF ALGEBRA. 19
13. 6x*y«4ar^y« + y^ x! + xyi + xiy + yl.
14. aH — ab^j x» + x\
15. X + a; 2/ + y* ; « + "i Oi + 0.
16. a^ ia+ 2; 2*5^*3«'y+ ixy\
17. a — aJ ; x* + x*y + x'^y" + x ly' + y^
18. X* + y^ + ;g'^ +  xy  X « — y « ; x'^ + .75.
19. x42y + z. 22. 2a^ — 3ab + ^b\
20. x*ia;§. 23. 6aj^yi.
21. X* — x'' y + X y* — y** ; x'— x* y + x^ y* — x' y* + xy* — y'.
24. a»2tt«i' + 3a*^»^2a'^° + ^«; a»+ 2a^i + 2 a6^ + ^»».
25. 2x** — 4xy" + 22^"; x^'^x'y" ^y'".
26. X* — y" + z' ; 3' + 2'.
27. a^ + ix»y4 ^% x'y^ + ^h^y'i xi'» + 2yi".
28. yx" + zx" + r.
29. xi + y*; x^ — 2Jxy + y'*.
30. xJyl + sri; x» + x''y + xy»+ y' ^^'
X — y
Exercise 27.
1. 7/1* + //? n f J/'* ; a* m* { a^ bm^n\ a^ b^ m^ n^ {■ ab^mn*
^ b*n*; m* n* + w^ n^ + m^ n^ ■\ mn \ \.
2. 1 + m n X + m' n* X* + 7w* 7i* x* + w* ;i* x* + m* 7i* x* ;
x«»y« 4 xV« + x«y 2* 4 x\//2* + xV^* + a;«yz»
4 x*2% 1 + ahX'\a''b'^2? + a''Vv+ a*6*x* + a'*6«x'»
+ rrV/«x«.
3. a«+rt»^ + ft*; x" + x'y' + y"; x" + x^V + x»y< + x«y«
4. a" + a»i« + a*i"+ aH^.+ i^; x""H x" 7/ + .r'*"?/"'
+ 2*x* + 2*x"' + x*.
20 ANSWERS TO THE
5. 16 a^ + 12 a^n^ +9 n^', 4:x^y^*'+ I.
6. a'^x^P + a*b^x^pf"' + aH^x'^y^"' + b^y^'"', 16x^+2Ax^y^
+ 36x'y' + 54.x'2/ + 81^1
7. xs + icK?/^ + x^^yi + yi\ xi + ic^/^ + »^^y + ^V^
+ x^iy^ + y}', a^ x?+ ai b^x^ y^s + a^ b^xyi+ai bx^yi + b^y^.
Exercise 28.
1. 12ba^x^— l^a^mnx'^ + ^5 am^ 71^ x— 21 m^ 71^', x^x^b^
+ x'b^ — b^; x'' — x^ + x^ — x^ + x^— x^ + X — 1.
2. £c5  2/?; 64 a;^  160 x^ + 400ic  1000; x^""  7/\
 1^5 2/^ ; a^aH + a' h'  aH'' ■\ cv' b^  a'' b'> + «' ^'
 a^ ^7 + «, ^8 _ ^9^
4. 243 ai<>  162 a^ b^ + 108 a« b^12 a^Z/^ 48 a2^»i2_ 32Z;i5;
ax" — 5'^ 2/2"*.
5. a^ — a^x^ + a 5 ic « — X ^ ; a^ x~ ^ — a^ ^s x%~i
\ ab^ X ^ y^ — b^ y 1
6. icl« — £c"2/T2'" + xi'*?/i'" — x^"2/i'"' + xi"?/3'" — ?/?5"';
Exercise 29.
+ ^22/2_.^iy3+ y4. 2o6x'192x'y + lUx'y'
10Sx7f + Sly\
2. 64xi« 96x1^7/2+ 144xi2y4_2i6x« 2/'+ 324x«2/'486xy^
+ 729//2; Slx^^5ix^7f + 36x'y'24.x^y'{16y\
3. Xl2n _^iou ySm __ ^8«y6m_ ^6« ^^9m _j_ ^4« ^12m _ ^2n ^15m
__ yl8 m . 7^18 a_jA6a ^^5 n __ /^12 « ^^^10 ri _ ;^9 a ^^^15 n ^ ^6 a^,^20 »
 A;8« ^25" + m»^" ; a^^  a^ ^' + a^ b^ — aH^ + a^ b^
 aH^ + aH^ aH' + aH^  ab^ + b^"".
4. 7^1 ^1 _ mi 71^ x^ 2/1 + m§ ?i? £C^ y^ — 7?^^ ti^ x^ ?/5 + x^ y^ ;
x^ — x^y^^xy^x'^y^ + y^', x'^x'^t/^ + x^ y""
 xiij^ + xUj^  x'^y'' + y^.
ELEMENTS OF ALGEBRA. 21
5. af X* — a» 6A xi y» + a^ b^i x^ y^ — a^ b^ xl if + a? 6i? x y>
6. a'^(iH'\b^; a;*a:Hl; a:«a;* + l; ai8a«6»4<^".
7. a;»x«/ + «*y*^'/ + /; a;''xy + y°; 164x^0;*.
8. 16a;*36a;«/ + 81y^ a;*^ «« + tV ^' b^ a;'+ ^i^;
9. ,,12 _ ,,« // __ /,ii 5 «i« _ ai4 ^2 4_ ai2 ^4_ ,jio ^e _. ^8 6«  a« ^»^«
4 '*' b'^  a^ b'* h b'' ; ,V a^'  3^ ^' f + iV y'
10. >(•'* — a» ' ^^• + b''* ; a«^ — rt^s ^,4 ^ a'" ^»8 — a*' ^" + a^« ^^^
 a" &» + «« 6'^*  «* />^8 + 6«^ ; 81 x*  9 a!^ + 1.
11. .r"'  ./•'» / + x""* i/''  x" i/^ + ^12 2/24 _ a.6 2^80 ^ yj6 . ^m
^12 _ an» + «• ^'  <^' b^ + ^''.
12. .r3«  x" //" + y»« ; x«  x^^/ + x«« //^^ _ ^so ^is ^ ^24 ^84
 x" 1/^ + x^ //«  x« //« 4 y/*8 ; yes ; a« _ «« ^4 _^ ^,8 .
a4 ^« _ ,^2 /,8 ^9 ,^12 ^ ,^^18 ^^24 . ^12^^20 _ ^9 ^16 ^,^6 ^2
+ a« 6»« m" /i^  «» 6« wi" 7i« + m" n*.
13. 2 + ./, 42a + a2; a» + i', </«^»«; 2  x, 4 + 2x+x'^;
x«+9, x«9; a*^ — 6*, a^« + ^/6^;<+ ^^8; 9««^4^»^
9«6_4^4. a2__25, aa25; a»J», ««+a«i« + i«.
14. x*y», x^+x^V+^V+aJ^iy^+.v"; ^w+x, 7«*7//«x
+ tTi'^x^—mx*\x*', x« + .y«, x^ — i/; x* + 1, x« — 1 :
a« + ^*, a« — &•; a*x*' + b*y*', a»x»' — (^« ?/'';
a; + 2,x*2x« + 4x«8x+16; 4a« + 9, 4a''*9.
15. 2a — 6, 16a* + 8rtV; + 4an«4 2«V>8 4i*; 9a< — 4/>,
9a* 4 4ft ; 1  y, 1 + y + y* + f\ y* + ?/ 4 ?/;
oa; + 10, a«x«10ax4. 100; a^x'^l, a^x'' 4 1;
a + 7WX. rf^ — r/'/nx 4 a^ m* x!^  am*x^ + m*x*j
X y — 9 ^, X y + 9 a.
22 ANSWERS TO THE
16. 2 a^ _ 3 h% 16 a' + 24 a^b^ + 36 aH^ + 54 aH^ + 81 b^"";
+ b^""; a^x^" + b^i/^"\ a^x^"" — a^ b'^x^^'y^''' + b^y^"",
cxP + b y% c^ x^P — c^ b x^^ y"" + c^ b^ x^^ y^ » — c^ b^ x^^y^ "
+ etc. *
17. a;2" + 2^2«, x'*" — a;2"2/2"+2/~^"; 2a;y+9, ^x^y"^
— 36 a; 2/ + 81 ; a^ f» — b^ /", a^ 2/«^ + ^;« 2/«« ; c' x''
+ ^>* ^Z"", c* x^^ — b^y^""', XT — 36, a;^ + 36 ; a^" — ^2«^
18. 2 a;8 + 3 3/'^ 64 a;i«  96 cc^^ ^/^ + 144 cc^^^/^  216 x^ y^
+ 324 x^y""  486 a^^ y^"" + 729 ?/" ; 16 a;« + 9 2/^, 16 ic«
— 92/^; a'"6" + ic^/, a^"'¥'' — d^'''h^''x''y'^a^'^b'^''x^'y'^'
— or b" x^'' y^' 4 x^^y*' ; l + 2x\\—2 x^ + 4 ic^ — 8 ic«
+ 16 a;« — 32 a;i<^ + 64 x^^; a"*" — 6«", a"*" + ^«»j
x^y^ + Ij x^yi — 1.
19. ic^r. ^1 _ ^^1, yj^ ^2„ ^1 _l_ ^^^i. a^jci+ 1, a2^2
— a^ici + a^x^ — a2ic4 + l; fa^ici^H ^'~^2/"^
— i 2/^ tV ^^" + ?^T ^"2/"' + T^o x~l^y^^
20. ^'s ci« + .3 £ci ?r ^ ^^ ^i'' — 3 ^> cf « ic^ ?/ ^ + .09 b^ c?« a;? 2/~^
.027 ^>i d«a;? ?r ^+. 0081 a^t 2^1; ^16a^icf " + .090!^"',
16a5a;l«.09a;J'"; 2t«aTV3^ i^,2§"«V<. + 3^^»i
Exercise 30.
1. 3 a^h^ — ^^ a + ^V" ^; ^~^ — ^^y~^ + .V"^
3. l — 2a — 2aM',xl\xh y\ J^ x^Tjh ■\ ?/l.
4. {a — b — c)"^ — (a — b — c)2'« — (a — b — c)"".
'5. 0* + y + ,*; ; a?^ _ 2 a' ?/ + t/^.
ELEMENTS OF ALGEBRA. 28
6. x* — 2x^f/z{ 4y^2; ar^ f 3xy + '^^xz + 3/ + z\
7. a;2y + a;ijri+y'.
8. 2 X*" — 4 x"y" + 2 y"^». 9. a;" y — af " * i/^\
10. a'^  2 a^b" + 6=^"; a^' — 1 ~ a^'.
11. 3a*'' + ^ — 4a«'' + 2a=^«^ — a«2.
12. 2al — 3a>*« — ai\
13. Sx^ 42f^ + ox''^ — af\
14. 2 m'^ + 3 m'^ — 4 ?/t'».
15. Saf — 4.af^ \ Saf^ — af^ 16. a"'"— ' — ^("^>"'.
17. 2a; + 1, 4a;''^ — 2a;+ 1; 4 + 9 a^ 4 — 9 w^; 4 ^  26,
16a''« + 8a6 + 4 6'^; a + 10, «2l()« + 100; a^ — S,
a* \ S] m — n, m* \ m^ n + w,^ n^ \ m n^ + n^ ;
l2y, 1 4.2y + 4/; aJ1, a«6«+ a«6^ + «*Z»^
+ ««6»+ aH^H a^> + 1.
18. a;=^+a:^a;2— a:2; WaX^'x^^— ^^^ul'^h^x^^ \ ^^aV'h^'^x^^
 iT.?«^''*"a:'''+2^?Z>^'; a:^" + y*"', a:^" — a;*«y^"'
+ y«; 05'  y^ x^« + a;^^ 5^ + a:^* y" + a:'y" + y«> ;
3.6m _ 2.8m y8H_,.y««
19. 2a« — 3y», 4 .r< f 6 a^y* + 9y«; 4 a* — 3 7i», 16a»
+ 12 a* n« + 9 w«; ti «*" ar^ + I ^*, 3 aA" + 2,
81o! — 54^A'' + SGrtJ" — 24aA" + 16; cx« — ay"',
c^a;*" 4 ac^x^^y"' + a^c'^x^y^''' + ««^«"y»"' + a*y";
J a^ + .04 6J% I a^" .04 ^J"; 8" a*" + 9", (64)" a^"
 (72) a*" + (27)*.
Exercise 31.
1. ± 5a;y^; ~2a26a;«; 5«i''; ±3a*i^
2. _7aW>«; ±ar^y*«*; a^V; i'^^ST'
3. a:*; ± 11 a;«.y; ±5aft; ±2^26^
4. — 3 a" ^.*;  4 7/1 n^ a«: 7»^ w«.
24 ANSWERS TO THE
5. la'y^', 2a3x2; ±^0"})^.
6. la^hc^'d^', ±^0,^1)"', 2;  2 a».
8. £c'"^; 2a2 6*a;«; iOir^^'^+^j _2cc"2/+s^
10. ± 2x''y«"^'^; — fm^Tii; a6^c\
11. \\ a bi c i, or — \% a b^ ci.
^ i_ 13 m — l
12. V84; 5''xf; 3»aH"; ai; xy, x » /; icy^; a^b^^c^\
{x + y) (x  yy.
13. (a h^ c")"; «'» {x — y") ; a'^ x^p ; (a? + ^)2«.
Exercise 32.
1. y — 1; 3 a^ _ 2 a — 1. 10. 1 — a ^ a'^ — a^ \ a\
2. 2a33a26— 5a62_^76^ '11. Smzz + '^^^r + y.
3. x^6x^ + 12x—S. 12. x^3x^y + 3xy^y^
4. a2+2a + 2. 13. 5a;2_ 3 ^^ _^ 4^2
5. 3 + 5 X — 2 x2 + x8. 14. x^3x^ + 4:X — 5.
6. ab — 2ac + 3bG. 15. 2 — 4 ai + 3 M.
7. 7 a^ _ 2 a  . 16. ^x^—xy\^y^; x^3x^2.
3. 2x + 3y — r)a. 17. 5 a:? — 3 £c* + 4.
9. m^~3am'' + 3a''m — a^ 18. ip^ _ 2^2 _ 3^1^
Exercise 33.
1. 182; 6.42; H; ^^; .315; 1.082.
2. .5555; 75416; 30709.
3. .2846; .9486; .0316; .3794; .5000; .0169; 1.8034;
4.5728.
ELEMENTS OF ALGEBRA. 25
Exercise 34.
1. a^^x—1; x^—ax—a^. 7. a — b — 2c.
2. 2x' + Aax3a\ 8. 1 — x \ x^ — x\
3. x^2x{l. 9. 2x^3a;y + 5y^.
4. 3a'»2ci6~^. 10. a + 26c.
5. l'x^lx3. 11. x"^ + a;y2/.
6. .3x^ — xi6. 12. 2y — 3x2/ + 4ar2.
Exercise 35.
1. 42; 32.4; .625. 2. ^^, or .0425; .0534.
3. .861; .430; 2.017; .669; .200; .873; i ^ii, or .637.
Exercise 36.
1. anbi'^crl; 200 a;» (* — .V") (« + JTY (^ — y + «")*•
2. — 32a\/2axy.
3. Sx^2 \xl
4. 2xi'' — 4 + 3a!ri".
5. 4 a;"' 4 2 aj*" — x'*".
6. a;iyJ — 2 4a;'y*.
7. aa:»2 6a!« + 3c.
8. i x«  2 X + i a.
9. rt" Ta;'"; a4±x*.
10. 2^^ + 4a;y 3a;«.
11. x^"" — 2a?^y" + 4y*". 26. 3 a* — 2 a + 1.
12. X* — 2ar> + l. 27. x^ + af — aj.
13. 2xiy«. 28. a 2.
14. 3aJ — i + 2aJ«. 29. (a + ft)^"* x + 2 a"r.
15. 1—3 a. 30. x" + x"* + x«*.
16. X— y. 31. 2 — a»"^
17.
2a + 1.
18.
±12; ±8.
19.
a 4 1.
20.
a^ab{b\
21.
517.
22.
384.
23.
a^ — 3 a + 5.
24.
X* — (m I w)
25.
5 a 2^ + 3
26 ANSWERS TO THE
Exercise 37.
1. 4 c. 18. a + 2b — IS G + 7Sd.
2. a — b + c. 19. — 6 a. 20. 0.
3. —Sx^ Sx. 21. 4 a2 _^ 4 ^2 _^ 4 c2 + 4 d\
4. —11 X  2 y, 22. 0.
5. 3 a  8 Z>  2 c. 23. 12 ^8.
6. — 35 « + 30 ^> — 30 c. 24. — 3 a 7i 2/.
7. 4 ^  • 16 ^ — 2 c. 25. 2 ny'^.
8. X h 2 y. 26. Saxh — 3m + 6 n.
9. 3 b. 27. 0.
10. 2xy 'J y — z. 28. 3m'^n^ ■\2 m^ n^ — n\
IX. ^^^a + S. 29. ^jy^ + ^xy.
12. a  J^ b Jr Y c 30 S x"^ — 8 y\
13. 3 a 4 4 :?. 31. a (^ + c) + 5 c.
14. 7x'hy. 32. (a + /*) — 9.
15. 210' Z> 222 a +84. 33. (a; + ?/) + ;^.
16. ^ xh. 34. (:*; + ?/) — z.
^^ ^^^CL' 35. (a + ^,)2_(a+&) + l.
Exercise 38.
1 «4 _ j^ ^3 _^ 5 ^2 _ 2] ; ^5 _ [6 w2 — 3 m8  3].
2. 3:r[22/5^ + 4?^]; »«/>« [2 a^ ^^ _^ a ^»«  Z.^].
3. A:X\3ax^—\Qx^ + Bcy~y'\\ x^ — y^—[z^—ab — 3acK
(3 or  2 y) + ( {4 7Z  5 ^}) ; (a^ b^  2 a^ b')
^(^_{ab^b'}). (4:x + 3ax^) (6x^{y5cy}y,
5.  [3ay  2 ab]  [5 b x  4. b z^  \2 c d  3],
[3a2/2f^^'4Z*,t]  [5^>ic + 2^cZ + 3].
ELEMENTS OF ALGEBRA. 27
6.  [ a + 26]  [rf  c«]  [1  ;.] _ [aj + 2y]
— [/i — 2 //t] — I4:abc — jj], — [2 6 — a — c z}
— ld{lz]  [a; + 2y2 w] [n + ^abcpl
7.  pxy2x]  [5a:«/4x^y^]  [xyzx'fj,
— [3 j:^ — 2 u; — 4 xy] — [5 ic^'y + xy 2 — x*y^].
3. _[_x* + 4«»][3a=^3a^][la],[4a»3a*x'^]
Pa'^a+lj;  [2 w4j»]  [37i + l][5x + 6y],
— [2 m} 371 — 4/)] — [5xf 1 h6y].
9. — [rtc — a/i] — [c'x ai] — [ax + ayj — [3aic — 3x^2],
— [« 6*— a ?t — a 6] — [c x\a x\abc\ — [ay+ 2a6c — Zxyz\.
10. (2a63«y46«)(56x[2c'(i3]). (a2^* + cz)
_.(2_</_l)(2 7;i — X — 2y)(7i — [— 4a6cj9]).
(2 X — 3 X y 4 4 X y^) — (5 x^ y — [x* y' — xy .^]).
(x» 3 a*  4a«)  (3^2 _[«_!]); (4 j9 2m 3/1)
— (5x— [— 1— 6y]). {an\ab—ac) — {cx\ax\\ay)
~{h^y~ [jixyz — 'dab c]).
11. m'hem*^?*! 12?^i7i + 8 7i» — 3m2x — 12mwx12 7i'*x
H 3 m x^ I G 71 x^ — x^
Exercise 39.
1.
9; 1.
3. 4; 2.
5.
1.
2.
A; i
4. 21; 25.
Exercise 40.
6.
1.
1.
16; 9.
4.
¥; 1 7. 4; ]^.
10.
3.
2.
12; 5.
5.
0. 8. 1%.
11.
2;6§.
3.
1; 60.
6.
Exercise 41.
12.
9
1.
1.7.
4.
66§; 20. 6. 2.
8.
8; 270.
2
3
5.
5. 7. 5; 9.
9.
t;i
3.
11.
10.
csj.
28
ANSWERS TO THE
2.
3.
4.
5.
7.
8.
9.
10.
12.
13.
14.
15.
17.
18.
19.
20.
22.
23.
41.
42.
43.
44.
45.
46.
47.
48.
Exercise 42.
12, 17. 24. 2 at 65 cts., 22 at 35 cts.
17, 31. 25. 25 lbs.
$20, $30, $40, $50, $60. 26. A, 60 ; B, 10.
Father, 48; son, 12. 27. A, 72; B, 24.
28. 7 years.
29. 40 miles.
30. A, 28 ; B, 14.
31. 103 gallons.
32. 3.7.
33. 20, 21, 22.
34. 24, 25.
35. 28, 29.
36. 100.
37. 200.
38. Watch, $117; chain, $68.
39. Linen, $0.32^; silk, $1.95.
1.
A, 25; B, 20.
A, 65; B, 40.
A, 25; B, 5.
8, 9, 10.
2,5.
5.
$50,
Father, 36 ; son, 12.
100, 200.
$23.40.
162.
First kind, 21 ; second, 42. 40. 11.
77 at 13 cts., 11 at 11 cts.
21 threecent pieces, 18 fivecent pieces.
Son, $1.04; father, $1.41.
76 tencent pieces, 19 twentyfivecent pieces.
$70, $52.
25 dollar pieces, 10 twentyfivecent pieces, 20 tencent
pieces.
Florins, 53; shillings, 71.
10 shillings, 20 halfcrowns, 5 crowns.
40 guineas, 52 halfcrowns.
ELEMENTS OF ALGEBRA. 29
49. 4 children, 20 women, 60 men.
51. Oats, 20; corn, 40; rye, 120; barley, 480.
52. $1225 at 7%, $1365 at 8%.
53. f 180 at 15%, $150 at 8%. 50. 30, 10, 223.
55. Saltpetre, 6; sulphur, 9; charcoal, 6. 54. $5070.
56. 51 women, 65 men, 150 children. 57. $3.75.
58. A, 47f miles ; B, 37 miles. 60. 10^ cents.
59. $313^ at 15%, $16§ at 8%. 62. 7.
61. Horse, $375; carriage, $300; harness, $75.
Exercise 43.
1. 2x2 X<iX ax (I Xbxbxbxx; 2x^XxXXXxXyXy',
[\X^XaXbxbxbxcXc\ 2 X 2 X 5 X « X ^ X <; X r Xc;
5x7xa;XiFXa;X2/XyX«X«X«X2X«X«;
^XlXaXbxbXxXX', 3x3x2x 2xaXbxbXxXxXx.
2. ^aHxA^a'b', 3x«y3 X 3xV; "d ab'^x'^ y^ X*d ab'^x'"' ^',
13a46i X 13ai"^>i.
3. 4aU*; 3alMx; 8ai6ic2; 5aift^x»y».
4. x\'X^x\x^xix^x^\ m'" wJ^.m?", ?Aii''.mi''?/ii''.ml'';
a;4 . xi . x\ a;i • jc* . a;i . jc» ; xl xl xi, jc* • a:* • xJ • xK
Exercise 44.
1. n{77i + 1); ab(4:a + ftc + 3); 3a*(a 4).
2. xiab\ c)', x*y^ (39 y» + 57 x"").
3. x«(5 a; 4 7); 12bxy^ {6bx  7 i^  S axy).
^ 2 aaf y z {462 at^^ — 5S9 s:r^ + 616ay2).
5. 4 a 6 (a  1 5 6* + 5 c + 2 a 6» a;* + 4 y  9 a« c ari) .
6. a;iy(2 — a6a;l + cajly«); 6a;J(a; + 2a;i — 3).
7. iac»(i<rl + an4a4&lcl); a"a:"(a* a"x'' + x2").
8. a''^"c*(ft"<r'"4 a'ft*" — a^^c").
30 ANSWERS TO THE
Exercise 45.
1. (x + 8) (x + 11); (x  3) (a;  4); (a^  8) {a*  12).
2. {x + S)(x + 27)', {bcn)(bclS).
3. (aH^ + 12) (a^ P + 2o) ; (« + 11 ^) (a  6 i).
4. (6^^8)(a^' + 3); (^^ + 4) (t/^HH); (aH5)(a«+12).
5. (ab]2c){ab—5c)', (a'' +10) (a^— 12); (n + .5) (71 + .3).
6. (^'^+762>)(a28^^^); (x+5)(xU); (x''\25a'"){x''12a').
7. (i:c4)(ajll); (m + i)(m + ); (oj + 2)(x 13).
8. (a& + 5)(a& + 26); (a5^>:z;) (a 15^»ir) ; (^/HSa;^)
{i/9x^)', {l + i)x) (l + 7x); (m7a){7riSa).
9. (a + 9x1/) (a  21 xy) ; {x + ij + 4.) (x + y + 1),
10. (1 5a&)(l8a6); (a  ^ + 2) (t^  ^  1)
11. (xy+2){xyb); {x + 21) {x + 21)', {x'12) {x^ll).
12. [(a + Z^)2 + 1] [(a + ^)2 + 8J ; {x^^'b){x''m).
13. (a + 3 ^>2 c) (a  13 b^ c) ; (ic« + a) (x«  b).
14. (ic + 5 2/) (a;  14 2/); {X + 1) (a;  I); (x^"  20)
(cc2«23).
15. (aj + 1) (oj  1) ; (x' + 21 C.2) (^2 _ 22 ^2),
16. (x 7/ + 11) (xy — 14) ; (a" a;^'" + 11 if) (»" :z;'" f 3 ?/«).
17. (a; ?/ — 11 a'^ 6'*) (a; y — 17 a''^*") ; (a;  ^) (cc  1).
18. (a;2»2/'"+17a'"6'")(£c2«^2«_^3^m^.„). (j^_^^)3m__7^4nj
[(aJ + ?/)'"• 14 a^"]; {n'+ .11) (^i^.l).
19. (o^ + f ) (a^  I) ; (:r + 2.1) (:r  .1); (a' + 1) (a^ + i).
21. (2 cc  2) (2 X  3) ; (3 a:  3) (3 a:  6) ; (2 :z; + 6 a)
{2x + 2 a).
22. (Sa + 4.b){3a + 6b); {4.x2a) (ix 3 a).
23. (5 x^"" + J a") (5 x'^'"  i «") ; [6 (a  b)^" + 13 (a — b)']
[6{ab)^"  11 (ab)^].
ELEMENTS OF ALGEBRA. 31
Exercise 46.
1. (4x+l)(x + 3); (2y+l)(2y3); {3a^\x') (ia^x^.
2. (3a:)(H4c); (4 a; + 3y) (2a;  7 y); (3aa:l)
(2aa:f 1).
3. (?/i'3)(8w» + 9); (3a ll)(r>a 1); (2a + 3Z>)
(3 a — 6) ; (2 m — n) (m  G n) ; (3 a; + 4) (a: + 1).
4. (8 + 9a)(38a); (a:+ 15) (15a:l) ; (443a;)(l2a;).
5. (3x~2y)(2a:5//); (4x3y)(2x+5y); (a:5)(15x2);
(12 x  7) (2 a; + 3) ; (a + 3) (11 a + 1).
6. (35x)(6~x); (3x + y)C2x3y)', (l + 7x)(5~3x).
7. (8x + i^) (3x  4y) ; (2 x2« + 7 y*") (3 x'^"  y^).
8. (x4y + 2w + 2/i)(2x + 2y + 7?i + 7i); (x + 4)(2x— 7).
9. {x + y^3a3b)(2x + 2yab)', (x+)(Ya:l).
10. [(xy)»2x"'yi«][ll(xy)»''xy5"]; (3a+l)(9al).
11. [2 a" + 3 (x — y)'""] [2 a" + 7 (x — y)""*].
Exercise 47.
1. ^m2H7i2)2; (;,i_.,i)2. (^ a''  W b cY', a^tt  2)^
2. (7 w»  10 7i'^)2; (9 x^ y  7 a')^
3. (7ii»  7*)^ (1  T) wi7i)»; x^ (x + 1)'*.
4. (« + /> + 8)2; (7/1 + 0)^
5. x*(2«»5xy)2; (19 a^»c 2 rf 77i7z)«; (11 7/i7i2_ IOje?)^
6. (15x2~y2)2; {2a"'b'''y.
7. n«(7 7w + 3^)^ (:^+ i)^
8 (s«*+ i*')'; c(aM3fc»)«.
9. {3x},yy', {m7i\iy.
10. (a2a + 3)*; (2x + 2yf 1)^.
11. (ai  il)«; (;/ii  1)^; mn (ml  niy.
12. (xJ + yi)«; {m ni  a)«; (2 xJ + 3 /t)'.
13. i(a^brr,rY {^,xl^)\
32 ANSWERS TO THE
Exercise 48.
1. (l7x)(l + 7x + ^Qx')', {2x9f)(4.x''+36xf + Sly');
(6 x''a)(86x' + 6ax + a").
2. {xy — ah) {x^y^ ^r cuhx^y"^ ^ a^h^x^y" \ ahx^if \x''y')',
\x _ 1) (a;6 + a;5 + cc* + a!^ + a;2 + ^ + 1) ;
(3 a  Z>)(81 a* + 21 aH + 9 a?h'' + 3 a ^>« + h'),
(a b"" — m) {a" b^ + aPm{ in'').
3. (6a7)(36a2 + 42a + 49); 8x{l3x){l + 8x + 9x'').
4. (a8  4 ^2) (^12 + 4 aH^ _^ 16 a« ^>* + 64 a^^*^ + 256 h^) ;
(9 x12y) (81 c«2 + 108 ic^ + 144 ^2) ; (^^i _ yi)
(a;* + x^y^ + ^^^/^ + x^y"" + y"').
5. 5ir2(3a;4)(9a;2 + 12a; + 16);
2 a6 (a  2) (a^ + 2 «» + 4 a2 + 8 a + 16);
(ari — 2/~^) (ic? + x^ y^ + 2/~^).
6. (abxy){a^b^{etG.); (4.a^5b) (16 a* {20 aH + 25 P);
(iC" — 2/"^) (^2« _j_ j;c«ym _j_ y2my
*
Exercise 49.
1. (2 a 4 1) (32 a^  16 a» + 8 a^  4 a + 1) ;
(l + ic)(la; + a;2ccHa;^); (ic'+2/') (^'ic^2/'+2/');
(^' + 2/') («'  :e'z/' + ^^^ 2/' — x^ y^ + y^).
2. (a 4 2) («« _ 2 a^ + 4 a^  8 «» + 16 a^  32 a + 64) ;
(a;2+ Oy) (cc*— 9x2^^+81 ?/') ; (4.t2+2/2) (16a;^4a;2^2_j_^^).
3. (ab \ x^y^) (a^b' — a^b^xy { aH'^x'^y'^ — abx'^y^ \x^y^);
(x^ + 4 y^) {x'  4 x^y^ + 16 2/4^).
(10 cc + 11 2/) (100 x^110xy{ 121 1/').
4. (x^ + y^) (£ci2  cc^ 2/' + 2/''), ^i' (^' + 2/') (^'  a;' ?/" + ?/')
(a;i2 _ x^ if + 2/1^); 5 x!" (3 ic + 4) (9 tc^ _ 12 a: + 16) ;
(a;8 __ ^8) (^.^16 _ ^8 ^^^8 _^ ^16^^
ELEMENTS OF ALGEBRA. 33
5. {X' f y') (X*  x^y' + x2y2 _ etc.);
{x^^y''Xx'^^xh/^y^) ; (xy^ab){x*y'abxY\etc.) ;
(a' I n (a"  a' i" + 6»«).
6. (a" 4 *") (a"  a" 4" + i**) ; (1 + x') (1  X* 4 «») ;
(af> + y2«) (x2»  x»y2"» + 2^"*); .
(ar» + y*) (xi  xiy^ + yl).
7. (a*" + 6»"') (a^"  a*" 6»'» + ^"'*); (2a^»c + 3 a;)
(16a*b*c*—2iaH'^c''x + S6a'b''c^x^—54:abcx^\Slx*)',
(4a + i^) (2o6a*  64a»*^ + 16aH*4:aH^ + 6«).
8. (4x'^49a«)(16a;*36a«a;2 + 81a*);
a «' + J *') (^ «*  ^'ff «'^' + tV ^*);
(a« + 6 c) (a*  4 aHc + 11 ^^^c^).
Exercise 50.
1. (ax + by) {ax  5y); (4x + 3y)(4:X  3y);
(5aa; + 7 by^) {5ax7by^).
2. (ar^ + 5^)(x+y)(a:y); («^ + 9y2)(x + 3y)(aj  3y);
(x* + y*)(ar^ 4 y^)(x + y)(x y), {x' + y)(a;* + f)
(x'^y){x^y).
3. (aH^ + 9a;V)(«'* + 3a:yi) {aibZxyi),
(l\10a*b'c){l10aH^c)', (4^8+ 3 6«) (4 a* 3^").
4. (3a'' + 2ar»)(3a"~2x^); (^ a + ^ 6) (^ a  ^ 6) ;
(a;J + y*)(a:iyi).
5. (a;«+y')(a;'4y)(x*y); (a4^+c4<^)(a+^crf);
(x— y + a) (X — y — a)
6. (a4a:—y)(a—a:+y); {ab\xy{l)(ab\xy—l)', 4ab.
7. (a 4 Z^ + 2) (a  ^») ; (a 4. 6) (a  * 4 2) ; 503000.
a 47a;(aj42y); 2805000.
9. 12 (a;  1) (2 a: 4 1) ; 1908 X 1370.
3
34 ANSWERS TO THE
10. (a^« + 1) (a^n + 1) (a« + 1) (a»  1) ; xy{Sx + y)
(9^2 — Sxy + f) {Sx — y) {^ x' + 3£cy + t/^) ;
h {a" + h^) (a + b)(ab)', {ah + bh) (ah  bh).
11. 6t (a + 4 a; — 6); (2 aj^ + 3 2/"^) (2 a;"^ — 3 5/1)
(4 a; — 6 a;*j^i + 9 y^) (4 x'^ + 6 a^y^ + 9 y^).
12. 150000; 2{ah\2hx){ah2hx)\ (5a4"+3i^>'«)(5a5"3i^>'»);
(a: + y) (a;  2/) (x^ — a; y + y^) {x'^ j^xy + y^.
Exercise 51.
1. (a ^ b + c) (a — b — c)^ (a \ b — y) {a — b — y),
(^ab + 2) {ba + 2).
2. {5x\b{3c){5x—b — 'dc)\ (a\x + y\z){a\x—y—z).
3. (2a;3y+9)(2a;3y9); (a; + 3) (a; + 4)(a;27a;12);
(2 a:  1) (a:  1) (2 a;2 + 3 a:  1).
4. (4a;2+a;^)(4a52a;+i); (3a+:r+42/l) (3«a;42/l).
5. (a; + 5/ — m + w) (a; — y — m — 7i) ; (a^ + ^^ + c^ 4 ^^)
(^2 _^ ^2 _ ^2 _ ^2^ . 4 ^2« (^n ^ J«>) (^« _ J«)^
6. (;s + 2a!— 32/)(^— 2a; + 3y); (2a + l— 2a;) (2al + 2a;).
7. (x + 3^ — ^) (a; — 2/ + «) (ic + .y + ^) (a; — y — «) ;
8. (c + (^3a + 2a;)(2a;3«— c— c^); {2x—Sy{4.z\bd)
{2xSy4.z^5d); {b + c + 2x) (2xb c).
9. (5a«+4a2+a;23)(a;2+4a2_5a843); (y^bb + Sbx^l)
(3,_5J_3^a;+l) ; a2„^^«._2)(a"2) (a2«_6)(a2' +2).
10. (3 a 4 * + ^"^ — y"*) (a;" — y*" — 3 a — J).
11. («» + a^^" + y  3 ;s) (a^ + xS" + 3 ;s; — 2 ?/2m).
12. (2aj + 3?/ — 67?. — 4^) (2a; — 3y + 6 71 — 4;?).
13. (a'' — 6« + c"* + ^^m) (^« _ j« _ c"* — A;2'»).
14. (2a + 3a; + 4?/ — 8«) (2a + 3a; — 4?/ + 8«).
15. (a + lb3c)(a + ^b + 3c); (a^+ ab" 3) {a''ab''\3).
ELEMENTS OF ALGEBRA. 36
Exercise 52.
1. (3a^ ^ 3 ab + 2b^) (3a^  Sab + 2b^', (a^ + 3a{9)
(a'Sa + d); (ix^ + 2 xy + y^){4x*2xy + y^).
2. (x' + xy + y^(x^xy + y^)(x*x^y^\y*);
(9 «' + 10 a a: + 4 x^) (9 a^  10 a'x^ + 4 x^) ;
(m* \ m7i {■ n^) (m* — mn + n^).
3. (2 X* 4 2xy + Sy") (2 x^ ~ 2 xy + Sy^) ; (a* + aH + b')
(a*  a^6 + 6*); (9 a" ^ 6 a + 4) (9 a«  6 a + 4).
4. (5a» + 7aUi + 46»)(5a»7ai6a44^»«); (aj + ^iyi+y)
(xiciyi + y) ; («» + a;3 yJ + y^) (a:»  xi yi + y»).
5. (4a*+4a^^>i + 36»)(4a*4a'*^>i + 3i»); (3tt^+2a62+76*)
(3a«  2a62 + 7^*); (^ + jo* + 1) (/>  p^ + 1)
(;?«+;? 4 1) ip" P + 1) (i?* i?* + 1).
6. (7aH4a6 + 96^)(7a'^4a* + 96*); (3a;H3a;yH5y*)
(3x^3xy^ + 5y0.
7. (m** + m + 1) (ttj'"*  m" + 1) ; (x^^ + 4x» + 16)
(x«"4x"+ 16).
8. (a 4 a* ** — ft) (a — a^b\ — b)\
(a«" + 2 a6'"  ft^"*) (a*" — 2 a" 6™ — i^'");
(5 m* + 2 m » — 4 71^) (5 m'*  2 m n  4 7i») .
Exercise 53.
1. (a + ft) (a + c) ; (a c + rf) (a c — 2 ft).
a. (a.ft)(m«); (aft)(4xy); a(a + l)(a=' + 1).
3. (2x~y)(3aft); (;, + ?) (r  3).
4. (x — y) (a — 2 ft — 4 c).
5. (a  ft) (5 a 4 5 ft  2) ; (2 X + y) (3 X  a).
6. (2xl)(x«42); (ax  1) (a»x»ax 1); (a;2y)
{m — n) ; (a 4 x) (4 x — a).
7. (x + my) (x — 4y) ; (a — x) (4 a — 4 x + 5).
36 ANSWERS TO THE
8. (a — c) (S a — b) ; {a \ b) (ax + b ^ { c).
9. (ox + Sy) (ax — bt/); (m ~ n) (n ~ p).
10. (m — n)(m + n p); (2^/ — 3£c) (3?/ + x) (3 i/  a:).
11. (c + 7) (3 a  7 6  5) ; (ic  2 2^) (:r  3 2/ + 3) ;
(a, _ 1) (;^2 _^ i^>^
Exercise 54.
1. (a\b + cy. 5. (ab + c df.
2. (a5c)2. 6. (3 0^2/ 4 ay.
3. («^ + 6  c)^ (oj + 2)^; (^ a  3 6  t)^
4. (a — 3 a:)^ 7. (m — ?i — ^ + a;)^.
Exercise 55.
1. 10 ((K" + 1) (af* _ 4); (^2 _^ £c + 1) (a;'  a; + 1);
12(xy+l)(xy^).
2. (a:  .5) (x  .06); (.^ + f) (« + ^); (3  a:) (2 + x).
3. 3m27i(m + ^)(^  m); 2 (2 a  1) (4.0^ + 2a +1);
(a^ + 9) (a + 3) (a  3) ; 6a;«(£c + 6) (a: + 2).
4. (xy + t'^) (^2/  t); C^^'^* + <^o) («'^  f);
\.(a + by^i]l(a^.by^l
5. (ax)(«'.a:4); 2(x'' + x^2){lx)(2 + xy a(ha'+l).
6. (a^3+ f ) (a^«^) ; (a+o,) (^^+1) ; ( a' 3 a«) ( a^ m^g a^).
7. (ma)(w7i); (a4.8)(al); (ab)(4.a U 2),
(la"" b^^ + a;^) (7 aH^"" + 3yi).
8. (17 + a) (12  a) ; (x'  if) (a;^"  i ).
9. (tw + w) (m — ?i— ^); «» (a;  1) (a:^ + 1) ;
a' (l + b)(lb)(lb + b^) (l + b + b^y
10. (20+a;) (19~a;) ; (a^l) (8a^l) (a^"'+ ^3'"+ ^2'"+ a+l) ;
l(x — yy^ + 5 ^2] [(a;  2/)^'' — 5 Z'^'"].
11. (3a;ll)(2a: + 7); 12(a:+7) (aj + 2); (x + y) (x + 7j5);
(f a: — J^??^7^) (fx ^ ?/).
FXEMENTS OP ALGEBRA. 37
12 (H^3^ar)(lrVa:); (2ar2/)(ar + 3.v2a).
13. (a" + b'x) (a^ + />";/); (a  (i) (a; + 2« + i) ;
(9x^ + liajy  y) {'^x' 2xy y«).
14. b (a* + ^^) (</« aH^ 4 0') ; (a; + y)»; (x^" + aZ» + ac)
(ar^" — w ^ + ^ c) ; (ni { n) (m* — ?/i w + /i' + 1).
15. (rt + ^— c— r/)(a/> + c— </); (a— y + a; + 2) (a— a + y+2).
16. {x ^ 0) {x\ ai b)] {x^" + 2 a) {x^" — a — b).
17. 2 (57/i + ii^ ± 1) (25r?iH 50//in + 25w^ if 5m qp 5w + l);
(4w + 70(4w+2m7i+77*2'), 3w(12w^+8mw + 37r).
18. (^c*af — a*) (/•'•(j^x" — 1); {7 p'' + 13;;^ + 11 q^)
{Ip' V6p n + 11 q^) ; 2 m {m'\ 3 7^, 2 w (3 t/i'H n^).
19. (m— 7i) (2m— 271 + 1) (27?i— 271— 1); (m + 7i) (2??i + 7i),
71 (m f n).
20. (a + a c + /> c) (« — a 6 — « c) ; (8 m* — 4 m ri + 9 n^)
(8mH4m7i+97i2) ; (r)a;2+3xy+4y^) (5a:23a;y+4y2).
21. x(3a; + 2i/)(2a; + 3y); ^/'(Sar + 4y) (2a;  3y).
22. (a:» — 3ic)(a:" — a^» — ac).
23. {m^^\n^){m±2n)\ (4m + 37i  3j9) (3w + 3/> — 2m).
24. (x»«"  c) (x"" + a + i); ' (X + y + 2)(a: + y  3);
(6ar» + 4 a; y 4 4 y*) (5x^4 a; y + 4^).
25. 3xV(3« + 2y) (a; — y); (m  37?.)(m + 2 ti ± 4).
26. m\n\{ab—xy^z)(,iv'b''^abxy\?^xyz\?>z^\x'%f)\
(9 a f aJ" b^"^ — 11 i'") (9 a" — «»" ii*" — 11 A"*) ;
(9 a*» + 3 ai^™ — 5 ^»*'") (9 a«" — 3 a» 6^*" — 5 i*'").
27. 2(3x2y)(3x2y±6); 2 (m + 37») (m27t 6a);
a«(a''x^+4n^) («*x*4a«7i2x^+16a*7i^x*), a^(ax}2n)
{ax—2n){a^x^—2anx\4?i^)(a^x^\2ftnx\4n'').
28 (x + 7y)(aj4y + 4); 2(1  3 a  2^) (//  x).
29. 7»77 (m + 7i) (m — ti)*; — a (a + m) (a* \ 2am \ 2m*).
38 ANSWERS TO THE
30. (3x — 4:y)(5x\4:y — 5a); (a — b) (a — c) {c ^ b).
31. (a h c)(c — a)(cd — l)(c^d^ + cd { 1); (mn ± 8)
(w^Ti^ qp 8mn + 64); 6 w^ (4m + 3 w) (w — 2 w);
(a a; — 3 Z> y) (a — i/) .
32. (m  2 7i) (m  3 w + 16) j a;^ (3 a;  1) (a;  1).
33. (m — n) {6 m^ + 5 mn + 7 n^).
34. 9?«.'(a2+ w^), 9m'(a + m)(w— a); (x+42/)(x — 42/ + 1),
(x  4 y) (a: + 4 2/ + 1 ) ; (a:  2 a: 2/ + 2) (a:  2 xy  3) .
Ten.
35. (36a:132/)(18ar + 29i/); m(m+ 1) (m1) (m^w^lO).
36. vi{m\n){m^ ■\mn\ n^){m^—mn\'nP)) {y\\){x—l)
ixy^l); {x\2)Hx2Y.
37. (?/i + 4)(m45)(m — l)(m — 2); (3 — w)(ww — w — 3).
38. (x» — ^ — a:")2; {x'^ ^ y^) {x^ — y\).
39. 7o2a:(a;2a)(2a:a); (x«+y«) (a:* + 2r*) (a^*2r*).
40. a:2(12a:88a'2/H2l2/); a:i(4arJ±3) (16a:q:12a:i+ 9).
41. (a:  y) (a; — 2 ?/) {x + ari^/i + 2/)(a: — ariyi + y) ;
(a; + 1) (x"* + 2) (a: f 4) (oT + 5).
42. (2a + 3^') (2a3/>')(ar2a)(a:2 + 2axf 40^^);
{m + 2n^p)(m + 2n —p) {p + m — n) {p—Di + 2w) ;
(a: + i) (x  i) (a:^ + ^V) {^"" + tV); («'" + *'")
(a*" + 5») (a*"  b^) (a*"'  b*''  6 a^"* ^»2«),
43. (x^ + 2/') (a:  2 I/) (ar^ + 2a^y 4 4 y^) {x^  x^y''^y')\
(a;"* + 1) (ar^"* + 4) (a:^"' _ ar« + 1) (a:*"' — 4 ar^*" f 16).
44. (^\n— p){m\p — n){m\n\p) (m — n — p).
45. 4 (a:"* + 2) (2/" + 4) (a:  2) (i/"  4).
46. (a:"* + 1) (a:" + 2) (x^"* + 4) {x^  2) (a;^"' _ a:*" + 1) ;
(2 a:"* + 3) (4 x""^ + 9) (2 a:"* 3) (a;"» 1) (a;2'" + a:"» + l).
ELEMENTS OF ALGEBRA. 39
Exercise 56.
4. 6a«*x*; 2axy. 13. x {x  y). 21. 4rzl.
5. Za'^^y^ !*• ^^^y + y''' 22. x + 2.
6. 6x^2/^z^ 15 4(a^). 23. x  3.
7. 2a:iyJ. 16 4;s(a:y). 24. m  n.
8. 6(« + ^) ^^ 2x3. 25. or 2.
9. ^y^^. 18. x^'Sy. 26. x^^e.
10. a:*(3x+2). 19. x  y. 27. a:"  5.
11. 3a«x12a». 20. x^  x. 28. 2a:"5.
12. x + l; a:" + 6; x + 3.
Exercise 57.
1. ar« _ 3 X 4 2. 9. ^^ (^ — 3).
2. a:^ — 2x + l. 10. 9m»(wil).
3. ar» + 2 X + t. 11. a:  y.
4. (x  1) (x  3). 12. mn{x^ 3).
5. x{x a). 13. x^  X — 1.
6. X — y. 14. 2 X — 5 y.
7. X* + 2 X + 3. 15. 2 n (m^ + 4 m y + 7 y'^).
8. 3 X + 2. 16. 2 m" x" (x«  1).
Exercise 58.
1. 2(x + y). 3. X42. 5. x"  2. 7. x2y
2. ar» f X + 1. 4. 2 X + 3. 6. 2 x^ + 5.
40 ANSWERS TO THE
Exercise
59.
1.
x'Vy.
17.
x''  1.
2.
xy.
18.
71 (n + x) (w — x).
3.
x1.
19.
x — 2m.
4.
x1.
20.
xyi^yY'
5.
x'd.
21.
ab.
6.
'6x' + 1.
22.
a {■ b + c.
7.
x^ + ^y + /.
23.
X" + 2,
8.
x+l.
24.
x2.
9.
x{x^ b).
25.
2{x + y).
10.
2x'^.
26.
7^2_^8x.+ l.
11.
a"  h\
27.
n + 2.
12.
ah.
28.
3m(y'+4f2y + S).
13.
3a:2« + 2m2.
29.
x''Sx\ 1.
14.
(m n)(x y).
30.
2(:r+l).
15.
x^ + A.
31.
x2y\
16.
a" + b"^.
Exercise 60.
1. S19axU/z^ 10. (nxy(n* + a^x'' + x^).
2. lUm^n^x'z^ 11. x^6x^ — 19x + S4:.
3. aea^^^cl 12. 105xy^x^y^).
4. 72 m^n^y^. 13. :r^ — 1.
5. 12aic3y4(a?23/2)2. 14. (3a:+ 2)(a^ + 2) (;r+ 3).
6. m'^n^ (x^ — y^). 15. {a ^ x){b \ x) {c + x).
7. 12 a:iy2 (a,2 _ y2>) j^g^ 3mn {x  yf {m  n).
8. (x216)(:z;2_25)(^_6), 17. („4.^)2(^2_^2).
9. x(x + 2y{x+l){x\'S). 18. ^i^l.
19. (a:^/)(x2_^2).
ELEMENTS OF ALGEBRA. 41
20. 3a«x(3a:a) (2a: + 3a) (a: + 5a).
21. (x + 4) (X + 3) (X + 1) (x  2).
22. (xy)(3x2y)(4x6y).
23. a*  1() 6*.
24. {a I ^) (w + w) (x + y).
25. (4 a;  5y) (2 JB  7 y) (x + y).
26. (x^ + y^) (x*  x^y*^ + /).
27. 20x2y(3x+l)(5x + l)(4xl).
28. (af ^ + c + d)(a + ^' — c — c?)(a4c — i— e/)(a + rf— Z» — c).
29. x* + xV + y*
30. 6x2(x + 7)(3xf 5)(3x2).
31. 12x(x" + 2)(2x« + l)(4x"7).
32. abc(m — n).
33. (a — b){b — c). 34. ale {a — x) (^» — x) (c — x).
Exercise 61.
1. 2x*4x«17x24xf 6.
2 and 3. x* + 5 x* + 5 x'^ — 5 x — 6.
4. (x2m)(x + m)(x* + m*)(3x2mx + m«).
5. 2xy*(30x» + 95 x*y + 68x»y» + 32x«y« + 24xy*  15y»).
6. X*  14 x« + 71 x**  154 X + 120.
7. (x23x + 2)(x* + 3x*8x« + 40x96).
8. ar* + 2x»9x^2x + 8.
9. 3(6x* + ar*~33x» + 43x«29x+ 12).
10. 6 X (x  1)* (.r + 1)«.
11. x*45x»45x* — 5x — 6.
12. 2x>2x'3x« + 3x*2x«3x'' + 2x + 3.
42 ANSWERS TO THE
Exercise 62.
1. (a  ft) (a 4 by (a^  4:b^) (a^  ab + b^).
2. x^" + 7 x^« — 10 x^'' — 70 a;2" + 9 a" + 63.
3. (a;" 4 4 2/"') (x" — 2y^) (x^" — 2a;"2/;« + S?/^'").
4. 2(x + 3) (2 X + 3) (x^  1) (x^ ^x^ + l)(x + 2).
5. xy{a^x)(by){2by)(2b'xi/).
6. a*"* — b^"". 7. x*" — 16 a**".
8. (ic" + c) (2 x" — 3 ^») (a;2» j a a;" — ^»2).
9. (ic + 2y (x' + 4) (X  2) (x ~ 3) (x^  16).
10. (X2« ~ a^) (aj2« _ ^2) (^2« _ ^2^ (^6« _ ^6^^
11. Sx — y, (3x — y) {x + yf {x — yy.
12. 3^2  2 < 3 ic^ y (3a;2  2 a^) (a;  9 a) (2 x + 5 y).
13. 2 a;" + 1, (2 x« + 1) (x^"  1) (9 x""^  4).
14. The expressions are prime to each other, (a* \ a^b^ f b*)
(a + by (a  by.
15. a;  5 ^>, 6 (a;  5 ft) (x^  9 ««).
16. x^'* — 7 a;" + 12, (a;2«  7 x" + 10) (a;^" — 7 a;" + 12).
17. c a; + *^ (c a; + ^^) (a^^  c^)
18. m^ + a; 1/, (w'^ \ x y) (4: x^ — 9 y^).
19. a^m ^ ^2n^ (^2m _. ^2 «^) (^2« _ 4 ^2m)^
20. a;^ + X 1/ + y^, (x* { x^ y^ + y*) (x^ — 4 y^).
21. 5a;2  1, (5a:2  l)2(4a;2 + 1) (5x' + x + 1),
22. a;" — y"*, (a;« — .?/"•) (a^^" + x^^i/^'" + y*^),
23. a;''  8 a;* + 50 a;2  a;  42.
24. (x'' + 7x^ 12) (a;2 + 0! + 3) (a;  2)^.
25. (2 a;* + 5 ^2 ^ 3) (4 x*  49).
26. x(6x^ 31 x* — 4x^ + Ux^ + 7x 10).
27. a^s + 3 a;^  23 .t^^  27 x^ + 166 a;  120.
28. 3 a;^ 4 2 a;^  3 a!  2.
ELEMENTS OF ALGEBRA. 43
Exercise 63.
2 2x — 3y
2bxy' 37/i«aj«/' 2x '
« . n(x* — y^) 2 X + 1
m X — J
^ 3m + 2 .. ^ y"» „ * + a? 3 + a
^ 3^;^^ ^(^y>^ ^ ° rf^' 2
3m — 4 3(w + w) a + i + c a — ft — c + x
4m — 3' m — n ' a—b\c^ x — a{b — c
3 x — b m^ + n^ x"*"^
4m*(l— x)' x+c" * m * b(a + b)
7n^n\x X — y x — 2 5x^41
X ' x + y ' a + 4' 9 x» — 4 X '
x—y—m c — d a — 2 b 2x^^ — 1
Exercise 64.
^ ^ + ^^ + nr3^' ^^^ 2x»x3 '
3. x« + 3ax + 3a''+^^; a; + 1 ^ "^ ^
X — 2 a' ' x*4ar12
4. (xy)«; x + m + 14^^— ^'
X  n
X — 10
5. 3x«4x + 5 + ,f— i^; x'^ + y*.
44 ANSWERS TO THE
Exercise 65.
2m IIP' ^ 2(^H1). 2m« w^
' ic + 1 ' m^ + ^^' 2^ — 2/
2. . ; — ii— : : ;;• 5. — \
6.
a + ic'
m — n^ a
d
m«
2y«
7/1+^ =
' a:Ha;y + /
a
2
mn^
m — 1
3.2. y2„
7. ^i 7 • 8.
(m + w)2' (m  /i)^
2/' (3 a;^  + //")
^,rn ^ ^,«y. ^ 2/'^» "• ic'" + 2/"
Exercise 66.
n — m n — m —(n—m) n — m h — a
' h — a a — h ^ — Q) — ay —(b — a)^ x — m—n
b — a — (b — a) b — a
, +7 x; etc.
2.
m+ii—x^ —{x — m—nY —(x—m—n)
m — a m \ a — x n — a — b
n — b^ m — b \ y '' m — b \ a
m — x m — X (m — a)(m — b)
n —y ' (jn — y) (n — z)"* (m — c)(n — x) (m — y)
2a: — y — 3 a — c + 3
4. ^ ■
5.
{a + b')(a — m) (b — 2 x)"* (a — c) (m — n) (x — y)
ab(m — x)
mnxy (a — b) (a — c) (b — c)
—^y{^ — y)
abc{a — b){a — c) (m — n) {x — y) {y — ;*;)
ELEMENTS OF ALGEBRA. 45
Exercise 67.
amny bm*y bmnx ab^y ^ c 2 b 5 a
bmnybmny^bmny bmny abc^ abc* abc
cm\cn am— an bn . Sa^bm 3a^m Sd'b — Sabn
abc * abc ^ abc 3abm Sabm Sabm
bmn
Sabm
10 m n\20n^ 10 m^ — 15 mn 15 m — 3n
30 m 7* ' 30 mn ' 30 mn '
a ;* 4 (x^l)(x^^) (g  1) (a* 4)
♦• (x2l)(x^4)' (x^l)(x^^)' (a;«l)(x^4)'
(m — 7iY (m + 2n) {m + n) m' (q  6) (a ^f ^^
2 (g + ^) (<^' + ^') 4 (g*  6^
a* 6* * a* 6* *
120 m f 30 10 n — 5 9m6
*• 16 (w  2) ' 15 (m  2) ' 15 (m  2) '
7 2a«y m«  n«
2 a; (x — y) (wt + n) ' 2 a; (x — y) (m + ») *
m (m' — m X f x^) n a (m + a;)
m« + x« ' m« + x« ' m» + x« *
• a^ + xV + y*" a;* + x»y« + y*' a:* + x«y^ + y**
10 ^ (^ • y)* 5(^' + y') 5y(x^ K.xy4y') «'  y*
• 5(x»2^* 5(x»y«)' 5(x»y«) ' 5 (x»  y») '
X" i //" x^yCx'^y*") {x^^ t y«")«
46 ANSWERS TO THE
a{b — a + x) (x — a — b)
(a \ b + x) (a — b \ x) (b — a \ x) (x — a — b)'
b (a + X — b) (x — a — b)
(a + b + x)(a — b + x) (b — a \ x) (x — a — b)^
x (a + X — b) (x — a — b)
(a \ b \ x) (a — b \ x) {b — a + x) {x — a — b)'
— a —b c
13.
(a c)(bc)' (a  c) (bc)' (a c) (b  c)
x—1 6—2x
15.
(a, _ 1) (a; _ 2) (a;  3) ' (x  1) (a;  2) (a:  3) '
93a: 4 a; 12
(a._l)(a;_2)(a;3)' (a;  1) (a;  2) (a; ~ 3) *
mx — am x^ — nx
{a — x){m — X) (n — x) (« — x) (m — x) (n — x)
ax — am
17.
(a — x) (m ~ x) (n — x)
a;22a;3  (2 f a;)
(1a;) (2a;) (3a;) (5a:) ' (1x) {2x) (Sx) (5x)
(a; + 3) (a; 3)3 (x'  4) (a;  3)^
^^ (x^ _ 4) (x^ _ 9) (a;  3) ' (a;2  4) {x^  9) (a;  3) '
• (■T^16)(a; + 3) 2 (a;  2 ) (x^  9)
(a:24)(a;«9>(a;3)' (a;«  4) >«  9) (a;  3) '
jB»"»4.3af" a;*"* — 1 a;^"* — 1
^^' (a:*'«l)(a;«'«+3)' (a;*'»l)(a;^'" + 3) ' (x*'"l)(a;'^'"+3) *
ELEMENTS OF ALGEBRA. 47
Exercise 68.
6a»16a15 . IS b^ c +1S b c^\ 9 a^c\ 9 ac^SaH+ Sab*
^' 36 a ' 72 abc
12 g» 4 28 a;' 27 . x* + y* Aa + b
^' Sx* ' xV ' 3b '
cp\bm — an Sam ■\ 2han \ Ibbn
abc 12 an
9 a + a' + 12 . 6 (n — m) ^^ a — b
4. n > rrz ' 6.
7.
3an triTi n
a«3a6c + 6» + c* . a^' + a'c'^*c«
a 6 c a^h^ c^
2a^ 2w^
13.
8.
4 n — w —
^=' + «^
mnx
9.
31 47
19
16x ' 42m
30 y
n
a«6 + ft«c
• ac*
14.
11
ar* — y^ ' a;* — m*a5
m + 4 2
m — 4 ' 4 ?«,* — m
m4 w, 2ax
abc '^"' m — n 8a;' — a*
1 . Sx m — 7
' ar» 9a; 4 20' ^^^^' 4m(m«3m + 2)*
4a — 66 5
"• 3(m«n«)' (a;2)«(aj + 3)*
16 0 ^^^
' {m { n + x) {m \ n  x) (m — n — x)'
17 ^^y* . 1 ^ 2m^
•'^ a;*/' x»y*' ^'* m« + n»*
g'+ 7 2 m* — 2ma;«
**• ar« + 6x + 8* ^ {m^^x^Y '
48 ANSWERS TO THE
21.
a;
27.
^' + ^' ; 2y.
la;«
22.
96 x^
28.
1
(3 + 2 x) (3  2 x)^
X1
23.
0.
3^2 _ ^2 _ ^2 _ ^2
1
29.
(^  a) (X  b) {X  0)
24.
25.
X + y
1.
30.
c — a — b
(ac)(bc)'
26.
1
31.
2x» . Sa + x
2x + l
x^ — 4 ' a + X
32.
1
0;
2x2
64
9a; + 44 33^ ^^
(rn^x){x^2)'
+ x«
Exercise 69.
1.
J3 ^ 3.m + «
8.
m {m — n).
' 10' 2/'" + "'
2.
8 4cie*
9.
3.
1 . 3a^
1
2
10.
x^ a^ y" b""
a^"^ x" V 2/''
2  X  ic^ ' X 
4.
aj+l . ^
11.
(a _ c)2  6^
x + 5' rr^ — mn
471^
a6c
5.
12.
1
6.
m
13.
a;«l
m — n
a:«+l
7.
m^ — n^ . X +
14.
^2n _ y2m
2 (a;2» + 2/'')
ELEMENTS OF ALGEBRA. 49
Exercise 70.
acmx 3 4 xy . xy
3(a^b)\ a*^2x^y\ 2 xy' + y'
b{a{b)' x*^3z»y^4x^y^Sxy^{^'
x + y . 1 ^ b — X
m*— 2m44 sb*— y" * a — x
2 a; — 1 g + 6^ — c
**2x — 3* ' a\ c — b'
"• »*  n*x^'+x* ' a^ + a' + l
12. 2a;V4a;V + 2xV.
Exercise 71.
a^b^ 10. 1. 13. 1.
a — 12 >! «rfi+i
(J a:  y)« ^ , a "* «
3. i ^lJ . 4. X + 6.
xy 1 1
5 (a + ft) (g  2 y) X x*
• (a I 1) (X + y) ' „ J
2a6 b a
7. 1. 8. 1. 9. 1. 1^ ^'2+ ,.
^ m5 77ttr,r^; 8^^« 17 ^^ . 18. lOa.
h^ J8 6y" ■*^ aP^c*
50 ANSWERS TO THE
Exercise 72.
03 + 6, a ■} ni ^ an { cnx ^ mnot^y — 3ic^
ic — 6 ' h — m^ cm + cnx^ m^ n^ •\ 2 in n^ x
h ^ m \ n ^ 7ri^ — h n ^ 2 m
a ' m — n a 7i ■\ b in m — n
x^ + 1 x'^U 1
3. —^ ; t; T^ . 5.
2x ' x^lU  2j(rnnp)' 2x^1
.. . r (a+x)(a 'x') x'+l . , x'x+l
1 + a;2 a;2 _ 3 ^ ^ 1 4
7.
8.
14a: ' ic^_4£c + 1' 3(1 4 a;)
* + 2/ . am{ adn _ £C — 2/
y ' bni + cn\bdn x + y
m \ n wn
9. a + a;; 1. 10. 2 11. •
{m — n)^ m \ n
Exercise 73.
^' ~8F' 16x''y^'' x'2/ ' 256x12*
160^"^^. (a; + yY . rn'ixijY
^' Slm^nf^' {xyY' n'ix^yy'
{x^  i/y . (a  bY . a''(a''15a^+75a125)
*• (m + nY ' (2c^ + 3 6)2' ic^^
(«_5)2 8 a;'^  36 a;^ y" + 54 g;B y^»  275/»" . /^y'+"
5. . . ,..; Tin » I ;
_ r ai m^ x^ 1 '
ELEMENTS OF ALGEBRA. 61
Exercise
74.
x^ Sm*n*
6ac»
«*
m
1.
^^*' X '
7x»
' 36**
2.
a»» ' b^ '
4m?i^
5aH*
3.
L*a*a'' a~* .
aixi
5.
x + 1^; a«
hi
3i/z' V^'
4.
aj + y . /
.^7 *
6.
a ft. 3
r*'
2V(a;y)' ^
Exercise
75.
1.
1 . xS
1 + x^' x + S'
5.
46; g.
2.
Ty^ — Axz
6.
]§. 8. 0.
10. H
6««7xy*
*; 0. 9. 8.
a
Sx*y2»
7.
11. f
3.
9 x«  y«« *
15.
(a  cy  b\
17. 1.
4.
aj + 5
16.
a:2 4 3 x + 3 
12.
":+"%.»;
^:i
+ K
13. m« + — 4 ^ + i; :l _ ::^ + !^^Jf _ '^,
n^ n"^ v}' y' ny"^ n'y n»
52 ANSWERS TO THE
18.
19.
\n inl \n inf \ a/ \ a^J \ic** y^l
(a^ ax x'^\
l?~b^'^ yV '■
5^(y'" + l)(y" + l)(2/" + l)(y"l),orx*»(2/^"+^)
20. ic. ^ "~ y ''® ^
2^ 28 (a: + 4) ^^' ^2/ * 3^ ^
36. ^±^ . 41. 3.
^c + ac * a6' 42. 1.
2b (by — ax)
22.
23.
(«,_,)(5_c)
24.^.
25. 1. 28. 0. 30. 1.
26. 2. 29. a;. 33. 3.
27. 0; (a«68)2. 34 2.
a:(a^ + l) . 1. 1 45. '^
^^' x' + ^x + r ' («^  ^0 (^  ^)
32. '^; ««^'^c^ 46. 4
o#.
ax (S ax — 5 by)
40.
16 aH^
43.
2{a + b + c)^ + a'' + b'' + c'',
44.
1
c (a — c) (b — c)
^1;
2 a
ELEMKNTS OF ALGEBRA. 58
Exercise 76.
1. 10; f. 4. 6; . 6. 2. 8. 4.
2. 20; 5. 5. i. 7. 2; 6. 9. 2; 14.
3. 2.
Exercise 77.
1. 1; 6. 5. 8; 0. a 4. 11. 3.
2. 1.3; 2. 6. 4. 9. 4. 12. .04; i.
3. 0; f 7. 1^. 10. %. 13. li; 3.
4. 2; 0.
Exercise 78.
1. n — ;— ; wv(l3a). 5. ab; — — r
2a 2ah ^ n *
^* 36 ' a + ft <?
1 ft I ^
3. 4m + 8n; T 8. 0, ; ft1.
'aft c
a • w»
4. j^. 9. i^ + 27^
Exercise 79.
^ c , ,, 13. H« + ft + 3).
3ft 2a»
2. — • 3. 4. ^^ 28«+5ft«* 12. 5.
5 ^(a6+c). *• ^* 16. —mn + mp{'np.l^. 0.
6. 0,; 17a. 7. 3(nl).
11. 2.
a
a* 2 oh
o
b — a* a + ft
6.  "^ . 3 rf» _ 2 c« 17. 36.
21 19. — To— TJ— •
12 c a
7. 0. • 18 ^•
1 n (n^ 4 m')
54 ANSWERS TO THE
Exercise 80.
2. 31. 3. 84. 4. 36. 6. 1^^ days.
abc
ab { be + ac
days. 8. 10 days.
2abc . 2abc . ^ 2abc
9 —r, 7T' A., —^ 7 days; B, , , ,
ab\ ac + bc ac + bc — ab "^ ab + bc — ac
2abc
days; C, ^j—, r davs.
^ ' ^ ab \ac — bc
10. 48 minutes. 12. 16 miles. 13. ^ miles.
b{G
15. 150 miles. 14. 742^ miles.
16. 56 hours; 84 and 70 miles, respectively.
acn . . acm .. _, abn
17.  — ; hours: A,  — ; miles; B, miles.
bn + cm bn{cm bn\cm
18. First kind, — ^^ ; second, — ^^ •
m — n m — n
19. ^^. 21. 69. 24. 160 miles.
2 m n {2 m \ n)
4: m^ ■\ 4: m n — n^
20. , — 2—^, =^^ days.
22. $1200 in 5 per cents; 2000 in 6 per cents ; sum, $3200.
$100 ^•m . ^100 b(nm) . ^
23. ^ in a%; — ^ ^ in c%;
am \ en — cm am {■ en — cm
$100 b n
sum,
am, •\ en — cm.
ab 1 , a(3w2_^3w + l)
25. miles an hour. 26. — ^^ ^!^ — —  .
b — ac (ti + 1)
ELEMENTS OF ALGEBRA. 55
Exercise 81.
(a: = 24, I « = H
i3, = 12. "• 13,= 7.
3. }^ = 2J a. J^ = 24. 12. j^ = l^
Exercise 82.
*• ■jy = 3. "• (y = 12. " U = 3.
_ (« = 3, (a; = 38J, ja; = l,
'■ ly = 2. 1y = 21J. • (y = 4.
6?
SI
1.
'•iy = 60. *• iy = 6. "•iy = S
ty = 12. '• y = 7. ly = 
,x = 19,
r X = i»,
1y = i
1.
Exercise 83.
x = t, . (1 = 21, „ f« = 13.
)y = i. 1y = i5. ty = ll
a. J'' = J' cr^fi 10. 1'' = ^'
(y = 2. (y = 6. )y = 5.
fx = 3, <x = 10, 11 j* = 4'
'•iy=12. '1^ = 11. "•1.'/ = 6.
4 (* = *• 8 i*= ^' 12 i' = 216,
'••ly = 2. "•ly = 12. ''•ty = 144.
56 ANSWERS TO THE
Exercise 84.
^•iy = .02. "• 1^ = 12. '^ 1^ = 12.
3. .P = 2' 15. r = 12, 27. j^ = 8,
b = 3. y= 6. (^ = 7.
fa,= 10, ra; = 7, p = 4^,
ly= 8. U = 5. ^" t2/ = 5i.
5. 5=^ = 25' 17. 1^ = 10' 29. r = f
«L=4: ^°{,=i5: ='='i,=io.
^•{, = 12. ^^•{, = 1. ^^■], = .08.
f* = ^' 22 jx = 20, (a; = i,
iy = 2. ""• 1^ = 21. ^*iy = 2.
jx = 60, 23. r = 2' 35. r^^
10
11
12. 1^=^' 24. r=^'
1 y = 4. * y = 7.
ELEMENTS OF ALGEBRA.
67
X =
2. r
(y
(y
X
y
X
y
X
y
X
\y =
6.
7.
(y =
rx =
\y =
X =
y =
i
§.
1,
2.
44,
6,
8.
12,
6.
am—bn
Exercise 85.
9. r = ^'
15.
x = h
11
12
13
<x = .
\y = l
{^ = h
' \y = l
(x = l
iy = h
(x = ie,
\y= 7.
<^ = h
\y = h
Exercise 86.
16. i^ = J'
(x = i,
17
18
19.
20.
21.
x=l,
1.
(y =
« = iWr,
J"
I
x =
y =
ar =
y =
an — bm
a^b^ '
nq — mr
I a* + aft + &•
Iq — mp '
6. 
^ a + 6
Ir ^ np
Iq — mp'
aft
l^= a + ft
be
7. 
\x=''^p:.
qrp^
ac
., P9'***
a«4ft»'
l^,vy
« + fli
ax^ — a
OiA + afti '
8. i
X = Sf
Oi — a*
*,6
1 — a Ox
1/ — *.
axb •\ abi
Ox — a*
58
ANSWERS TO THE
9.
10.
11.
12.
13. <
14. <
am^n
h
+ b
a
a — b
,2
a'
{a — c)
y = 
a — c
nr
X — — ,
c
X =
c{a + b)
2^
c (a — b)
2a
2n
m \ n
2m
m ■\ n
15.
f m — n
X =
m\ n
aai (a — ai)
16. ^ " r «i
I _ aai(a + ai)
17.
18.
19.
20.
21.
22.
23.
r
\y
X = m { Uf
m + n.
a
be
I
a + 2b
x = ~ }
m
n
1
— >
n
L^ m
r a; = a  5,
\y = a~b.
(x = m + nf
\y = m ~ n.
a'^jb a)
^ a« + 6» '
y =
a^ + 2b^ab
b — a
24. {
X = zfy
mn — 1
y =
41
mn
25.
( X := m — Jlf
\y = n — m.
ELEMENTS OF ALGEBRA.
59
Exercise 87.
(
x = 3,
x = l,
1
x = ,
a
"1
(
y=2,
.z=b,
x = 10,
4. ^
f
y = 2,
z = 3.
x = 5,
7. <
1
1
"•
y= 2,
^•
y = 5,
'x = 20,
1
.2=3.
I
« =5.
8. <
. « = 30.
ar = 2,
(
:r= 7,
rx = 12,
'■{
y = 3,
.« = 4.
M
.;s= 9.
9. ^
y= 6,
U = 18.
a^i
5
>
2
^a6 + *c
4ac
~ ?/l + 71
10. <
abi
?
9
12. <
2
^~a6 + 6c
4 ac
^ mj p
abt
5
•
2
[ a4 + ic
+ ac
. ~ w +/>"
par = 4,
rx = li,
r' = i,
14. .
y = 5,
17. ^
;y = 2i,
11. .
[v = 3.
\x = a,
U = 6^.
a
15..
y = b,
18. .
y = 3i,
» = 2'
z = c.
U = 2^.
13. V
/=2
16. .
rx = l,
19. i
fx = 7,
y=9,
U = 3.
60
ANSWERS TO THE
2.
3.
iy =
3,
2.
13,
5.
y =
a + b
0.
4.
r
z =
a,
b.
6.
Exercise 88.
(x = 2,
13/ = 3.
^ = ^^^4752'
a^bc
\ X = 5^7^,
^ ~ abi + aib
2 a aib
^ ~~ abi + aib
11.
10.
mn
12.
13.
14.
^ — ^^
2/
U
— 3 5
y =
m
X =
15. ^ 2/ =
z =
f._
m
+ n
Sp
w
a
— n
n
b
+ 3p
— m
m
c
20.
f _ (a + ^) w +
_ (a + ^>) ^ +
2,
3,
4,
5.
^m
"^X"
16.
VJ.\y =
2 m '
mn \ mp {■ n p — n^ — 2 p"^
z =
2(n'
3np — mp 
p')
m n
2 (n^  P^)
19.
y =
z =
18. f =
ELEMENTS OF ALGEBRA.
61
X =
21. ^ V =
mnp
mn + np — mp
2mnp
z =
nip + np — mn
2mnp
VI n f itip — np
ar = ^a + * + c) — a,
32.
X
=
2
mnp
mn f
np
mp
y
—
2
mnp
mp +
np
mn
2
mnp
mn \ mp — np
1
x =  t
a
23
rar=f (a + ft + c)a, i
I z = § (a 4 * + c)  c. 1
27. f^ = ^^* +
W* I 7i*,
>8
30.
31.
34.
X = 105,
y = 210,
z =420.
a; = l,
y = 2,
« = 3.
x= 7,
y = 10,
2= 3
= 7W,*
35 (x = m''n",
2. J.
3 h h
4. ^.
Exercise 89.
5 A
6.
a\bm b + an
mn— l' mn— 1
10, 1.
7. 16i, 15.
cm — an
10. Corn, i; oats, f .
,, /t bn — dm
11. Corn, ; oats, .
oc — ad 'be — ad
12. 180 lit 2 for 3 cents, 300 at 5 for 8 cents.
62 ANSWERS TO THE
100 a (b c d  12 en  12 np)
. 777 r at a eggs for m cents,
13. ^ d(b m — a n) °° ^
100b(127nc + 12pm — acd)
TT, — c at b eggs for n cents.
a {b m — an) °°
15. 245. ^l^L±_?!l!!)
^^* " 2 a  81 ■ 20. 853.
16. 891. , ,, ,
19 ^^^ + M11^) , 21. 315.
18. 39. 11 — a — c
23. A, 120 ; B, 80 ; C, 40 ; altogether, 21^9^.
. ac(n — m) ^. ac(n — m)
24. A, ^^ ; B, ^ .
nc — a mc — a
25. Arithmetics, 54 ; algebras, 36.
26. Crowns, 21 ; guineas, 63.
,., nia^c — a Ci) . n(aic — a Ci)
27. Crowns, — ^ : guineas, — ^^^ ^ .
a^n — am cm — c^n
28. A, 50; B, 21f. 30. Stream, 2; A, 10.
. mn ^ mn
29. A, ; B,
31. Stream,
a a \ n — m
mhi — rrixh . mhi \ m^ h
2hlH '^hK
32. 39 miles, 8 miles an hour. 33. 5.
34. Going, 4^; coming, 1\\ stream, 1\.
hh . ah ^ m (a" — b")
35. Gomg, ^p^; coming, ^^; stream, \^^^ ^
m{a\ by
2abh ' 36. Sugar, 5; tea, 60.
r^ 100 nOO (71  m) hm']
3,J^^' TW^) ^
100 [am 100 (yim)]
^ea, ^^ (a  6)
ELEMENTS OF ALGEBRA. 63
38. Sum, $1000; rate per cent, 6.
39. Sum, dollars; rate, — r— •
40. First kind, 14; second, 15; third, 25.
41. Forewheel, 4; hind wheel, 5.
42. Forewheel, ; (rnr'sn) ^ ^^^^ hindwheel,
' (m + 7i) (cs + cr {• ar)
b(mr 8n) ^^^^
\$ + r) {em + en \ an)
„. , . . . bnti — bim aim — ami
43. First kind, — { \— ; second, — r j .
aib — abi Oib — abi
44. First, 3} ; second, 3.
^. ^ a(m\l) . a (n \ 1)
45. First, — ^^ T^; second, — ^ :r *
mn — 1 mn — 1
46. 3^8, 80% ; 4's, 125%.
bd \ np bd — mp
4a A, 55; B, 105.
A ^{n — b) (m \ n — a) c {a — n) (m \ n — b)
*^' ' m~(^r:^) ' ^» 7^7^^^6)
50. Sum, 500; rate, .04. *
., o bm — an ^ n^ m
51. Sum, —7 ; rate,
b — a bill — an
52. Pounds, 60; cost, 28.
53. Larger, 5.678; smaller, 1.234.
54. Smaller, :j ; larger, 
1 — no 1 — a
64 ANSWERS TO THE
1. n^^ i.
216 a« 1
3.
^ a.^ + 1
4 ™n + I . 1 . .
Exercise 90.
a' 4 /^«'*'Y
"•''*' u + w
^ ^
2" ' a' (a'"  b*")
11. ^"'f'.^ + lZ;^'''"*!;
1
1 16
16" a"""* a;""' y^
' • m m > 1 27 •
Exercise 91.
13. aVa;^ 6^2 6^c2«'.
14. 2". 15. 22'".
16. (a;2y2^3«j 2«.
1. ^ ; 1 ; ^a;^ 1. 8. V{a2 x)
9 ^\17(l + a)
a»' _i_. L_. 1 9. a;2«; 1.
1  „ 1 10 (^ + 2/r; ^"^'^ 6
3 ^r;;r; va»; Vet'"; 
«
11.
_1_ /^\ {m + n)^
4. — 4 a/2 ; ^a\ xy.
6. 3"; (a 5)"; g 13. ^a ; 2«'.
7 ^\'/"^. ^' 14 a a;« + ^+^"
ELEMENTS OF ALGEBRA. 66
Exercise 92.
1. 9; X; 4m*; ^; ^.
2. IG; o^^i'^; 516a»''6»»^c»''«.
3. 343 aV"*"*; (27)'»6»'« Va*"'; a:*
5. ,; y»^^; 25. ^ S ' 1 5 V^'
7. _^L^j (25)'"'»a'''r***'"c'^'"^"S J^4. 8. a:"'"; a^'^"
Exercise 93.
1 ,1 ,1 (b\
^ (^Y <""">' (^»' (''^^'' (;^' i^'i
Va'ft'y*''
_1_ « m
4. fo^; ^iO^^ yaHW', yJ a^^ b" ; \ ab^, ^ x^ i/\
6. Sa^bc^', 5a^b hr^; mlnij ab^; x^t/^; ^"*y^; «'**•
2 • 1 1 1 5 _ 1
^ 5a«ca:^ aJ^!^ ^^^ xjri^*^ Sa^J^V^ ^""^'
rt* 1
8. «*; jgj «*"•; a»; ; al.
66 ANSWERS TO THE
m
10. ; 7^TT^', 1 20. t; 1; 1.
12. ;;16.^m;^; 22. 2«%^
M , ^^ 23. I; 18X10^^'
\/nr J/7. ^"^ 3'
2 »i m t t
1 6/
a; "
Veto * "? ^ ± 1
27. a'* — a2»^»4« ^ 523
a^
Z.2 ^ 29. £c2n2_ 2m
17. ao''; fw7^ic. "^
t m t
11 . .
32. a** — 1 + a~'* ; n^ y^ — n^m^'xiy^ + mTn*xiy^ — m^n^x^y
+ m^n^x^y^ — m^nrxy^ + ?7i7a3s.
(as" + &'*)(ai'* — b'').
b^ Ab^x 6a;2 Ax x^ , ^ 12 8
34. ^ + — T 72 + 745 x^ — 6x + — + ^',
a* a^ a^ ab^ b^ x x^
5n . 5n
a 2 _ 5 ^^n _^ ^Q ^^n _ ^Q ^^n _,_ 5 ^f ,
a
35. 2! X 3^, 2i X 3^, 21 X 3^, 21; 24.
36. 2§ X 3^, 2i X 3^, 2i X 3^, 2i, 2i X 3^, 2§ x 3*; 576.
ELEMENTS OF ALGEBRA. 67
37.
ai + i; 26i
aH\.
43.
a''^+a; 2^^'
38.
2a\a\.
44.
77» ^
39.
x^x\
y'
1
45.
xi;2.
40.
X
,,^ ,; 2»«% 2.
X
46.
♦1.
32a^.
i^m"
42.
x^ . 1.
47.
(5a:^3yY; m« 
w«.
Exercise 94.
1. V^i; \^V"; ^64; ^i*""; ^8"^; ^27 aH*c^; '^/W^',
^; ^^; ^i^^. y/i; y/J; ^/2^; ^F;
a. Vi; ^i; V800; VH;_v/^¥^; v^^^.
5. VW^^l^^, v^ii^^l; V/^
8. 12 V2; 15\/6; 9v^;  \/58; \a</Va.
9. V2; f\^4; ^^^20; 7^3; iV5;iVl5; i ^32.
10. i'^'24; VV6; 3a6^4^; 2aJ6«'^; ;r^,\/3a*na;.
11. a*iV^c; V^; — V^^aJ.
68 ANSWERS TO THE
12. ^ ^6 ; 9 a" b^"* \/Wl? ;  ^2ax'y'; x y"" ^yx\
13. (x + y)Vx — i/; (x — 4:)^a.
14.
t>(^ + y)''''
7
^3n^2,n^5^
— m
15.
72; 3; 2.
20.
fa^c; a^oTx.
16.
17.
21.
22.
1.
3 2
2' «•
18.
10 a m X.
Bab.
23.
24.
2 a.
19.
a</4; 2a«Z»'^8a2i»
Exercise 95.
1. v^8; ^9; §^1296; 3^64; ^^»; ^8; J ^16.
2. V3; tv^64; {^«^'; •'^?.
3. v/a^?"; y/o^^; y/a^ SJx'yl', ;/a; y/^; :.'Y«'
4. v^l25, V^m, V^l3; v^i024, v'625, y'MS; ^^Gi,
^81, ^6.
5. ^2, ^2, ^^2; ^49, ^625, ^2l6; ^a«, 3^^°.
6. ^^, ^^; ^a^, ^^^ ^^; ^^, ^^^
7. V< ^625^, ^27^; VW^', V27 b^ V^^x^
8. a:^< ^^«"^; ^< ^< ^^, ^^^^^^^.
9. '^a', W, ^Z^; 4^625^«, 2 v'8T^^ 10 a ^729^
ELEMENTS OF ALGEBRA. 69
Exercise 96.
1. 2Vi4; eVH; 6^4. 3. V^l v^4; v/f.
2. 10 VS; ^ViO; ^2. 4. ^Ti; 1.6; V^.
5. V3, ^7, ^4; 8 a/2, 5 a/5, 4V7.
6. 4^/7, S'C^iO, 2V^21; 4v'ii, 2^l3i, 3^.
7. 3V2, 3^5^, t>^; 3\/8, 2^/8, 2^21.
Exercise 97.
1. 8\/5; tVlO. 7. c (6 c — a) \/2 a * c.
2. 2V3; 6^6;  v^2. 8. mnx\/mn^.
3. a/3; 17^. 9. 2bVS~a.
4. 20 J a/3; fA/15. 10.  ^'^^~J\
5. ^a/IO; 0. 11. fA/6^14.
6. — Vva*; 2v5. 12. — ^^ s ^a/wx.
Exercise 98.
1. 18; 18 a/7; 25 a/5. 6. ^^^486; v^ll8098.
2. 2; 14 a/G; i >^12. 7. § >^ ; J a!^ ; 14 >J^.
3. 9v^288; 10 (3 a/3); S. 8. 729;
,— 6(l3A/3 + 3ViO).
4. 2+ A/6 + 2VIO; 120. ^
5. vTO; T»B ^2592; ^500. 9. A/ lOO; nV^; 12v^2.
70 ANSWERS TO THE
10. in n ^72li^x\ 19. 5 ; 4 \/32.
11. m^m^7i^x^^', 6\/^iH\ 20. 4; '^3.
12. v'72^; 2«m"a;". 21. \/^ Vn, yVv^lSOOO.
13. ^V2; 121 V^. 22. 2; v^9^^
tt X
14. 9ViO12 ^32 + 6^^31258^/10; ^^4 + 2 ^6 + v^9.
15. 4(6VlOl).
16. ^^128  ^^2187  '^972 + v^O + v'i  V6 ; 12 ^^27.
17.  1 ; 'v/y  2 ^^2. 23. 49 a;  9 a.
18. 2 ;  7. 24. 1 v^288 ;  ^^80 ; 20 \/b.
Exercise 99.
1. 2(a/3V2); ^5(2^/5+^6); 3^/24; 3(VT5VtO).
2. 4+V2; l^yV^ ; _^(3 + 2V6); t^re.
x—2'^xy\y '\/xy{?»x\^\/xy — 3y)^ a\^d^—x^
x — y ' y{x — ^y) ' a;
4. 2 i«^  2 X V^^^^l  1 ; ^V^(^^ h) { a  Vb)
a^b
5V5  VTO + 5^2 + v/8 2 a \/a:' '"«'«
23 ' 3 a; s
5. 14.14; 7.07; 141.4; 11.18; .1732; .2236.
6. .707; .236; 1.266; 1.216§; .268.
ELEMENTS OF ALGEBRA. 71
Exercise 100.
1. 3V3; V2; '^3. ^^ axV^^Vy; — ^^^^^.
3. 3.3; ^VO; i ^486. "^ ""
4. 2^3; ^54; ^18. ''• V ^^^^ ^ ;  ^1944 aV^
5. 6^5; i^96; 2^/2. 13 ^^'^^^^5 1 4 V^.
7. 5 V7  4 VC + 2 V5. ^3 5  V6; ^ '^a^^^
. le/^
3aa;
8. .2; a^6c; ^aft*. ^ .^^ ^ , /^
* 16. 1 (4+ V15); r)+ V<.
17. ^x + V^^ + Vy.
9. !!Lzl!^; 1^96
Exercise 101.
1. ^^; V3; 2V3; V^8; 2 A^^i
2. i>^; i^J^; V«^; 2.
3. 4^; ^n»; 2^'/2; V2; 2 15^2.
4. V^; 3^; H; 4«^3; ^ ^a.'
5. 8aVft; ^V^i^^^^; v'^; 32a'bc\/2abc,
6. 3a; 648^/^a:^ 2a«; ^^
7. r,; ^(«rr; 1; ^^^^
4
aC
72
ANSWERS TO THE
8. JV3^; a'"V«; va''; 4
1 X,
9. \^n^—''; "''^(aH'^)"^^''^^^; m'n' ^/n\
n
1 8/r 0^^ iC*' V^^^ ^^
10. ,; Va<o; — ^g
11.
^.; ^2/(^ + 2/)^ 2a^^3.
27
12. x^ + 2;r27/2 + 2/'; t + 4V6; 11 V2 + 9 a/3.
13. 2  3 ^y + 3 v^3 ; 4 + 2 V2 + 6 ^4 + 6 '>^32 ;
m'
'\/')
m^
3/— , O 6/
iiin
14.
_^ ^6 — 4a" + a^"; v<*^ — —
^58m 16
15.
5 10 10 5
^12 + a« + a^
^^ + ^e + ;;r2+:;F+4 + l; «* + 4 a« 5 + 6 a^ 6^
16. ^V2m7i + 2^m7i2 + ^^8m7iH404 — ^Sm^Ti
, 32m 3,—, ^ , ...  ,
Q>A:m^x^''.
Tn^w
w
17. ^ + 13^2/.
19. ^ + «^^2/'
2 2/^'
Y2
18. 1 — 2" ; 3" — 2\
1
6"/;:;57
20. ^^a^2V;i^ + V^
ELEMENTS OF ALGEBRA.
73
Exercise 102.
1. ^9; 1.2; 9; 1; i; .
2. a»v^2; mv^G; i^768; 4.
3. ^^5^; ^{/243; 6^3', 4 Vm (m + n).
4. V^^?^^; i^xTy; e^^^^v^a.
5. ^64, ^?^, ^l25; ^6561, '^EV^, i?^15625; 15^2?;
9^4.
6. 6^*64, 6^16, 12 ^8, 10^4; ^a\ ^6", ^^\
7. Va, 2V3; 3v^2, 3^2; 2v^9, 3^9.
8. iV5, a/5; V^5,3a!^5; Vtr VlO, t^^VIO.
9. §a^9y^2ftv^96■^^^^'9P; ^ ^50, ^ ^50.
10. l</% \^% ? ^9, ^; 2 V5, ^ V5, 20^/5.
11. V2, V2; 2a;V2, %bW% « V2.
12. i ^, 2^6. i ^6, i ^6, i ^6; 2 v^, ^2.
13. 4^^, 3 \^3, 3^/275, 3V19, 5^8, 3^9.
14. 2 ^8, 5 ^2, § V3, i V5; ^H, Vf
"• ^^^VS^» ^^5^(2i^^,aV(8^^.
16. 16 V3; 7^7. 20. 8^4; v^3.
17. 4>^l6l25; 3 15^4; ^624288.
18. 3^/2;  ^3. 21. H\/15; T^y.
19. 24v^4; 1. 22. i; 6.
74 ANSWERS TO THE
23. 172; 0. 34. 1; i; ^^ ^^.
24. /s'^^; ly^. 35. ^^V^; 0.
25. 15624V4. ^^ cm
_ 36. 11; — .
26. f(72VlO);iVl5. ^"^
Vw/^ + «^ ^^ — <^ ^ 1
27. ^l:::_lj^l^! — ^. V^«z/ _j V2+iV3+^V3o.
77t y
28. ^ v^lOSOOOO ; 204.8; ^a'h'',
29. 47; 27. V^. icB  2
30. a^i + ljar^yt^n+D.j/^ ^ ' xl + 2
31. a^^. 39. V2.
32. cVc; ^a. «,y^2o X 315
,.y/^
33. c ^c ; h '^Wb, V 2^^
38. 1 ; Va (a V^ + 3 a + 3 \/a + 1).
Exercise 103.
1. 3V^^; 2\/^; 2V3^xV^; 2aV^; iV^.
2. 7a"^»V=^; 3'^=!; a^^/=l; 2'^^.
3. 1; V=^; a/=3; +1.
4. +V^; +1; V=^; V^=^.
5. _4Vin[. 8. 2&(2a2jl) V^.
6. 4 V^  i V^3  3 V^^ ; 0.
7. a2(8a + a2_^y'ZrX. 9. 3V2; 36^/3.
ELEMENTS OP ALGEBRA. 76
10. 48; 2,^/^i, 16. 6»a»
11. ~GV<i — ^9. 17. V3j AV3.
12. 392V^; 21. 18. i; 2V2\/3.
13. 2; — 4 — 5\/^^ 19. 4;  Va — V^.
14. 30 + 12 V2. 20. 1  V^; 1  V^.
15. (a  h) V^^; a + «. 21. i A^2625; V3.
22. 1 + V^^; \ V^^  i a/3.
6 + 2V^ + 3\/^~a/B 24 + 10V315V68V2
^ 3^6 ^ 43 5
lV1.
^ / — T a* — 2 a V— « — a; 2 \/a^ — a — x
24. 1 — V — 1 1 7. : } •
Exercise 104.
1. V5A/2; V3 + 'v/2; 4\/23; iVB + 1.
2. VI02V2; V6 + V5; V7  \/6.
3. V141; 2VllV'3; V^l; V6 + 2.
4. 2iV3; V^2 [a/31]; a/5[V2 + 1].
6. m Vw — V^ w; a/'J'  a/2.
7. m — n — 2Vmn; /v/^('y/2 — 1).
8. 2V3; i(V5+l); a/5 + 1.
9. a/2 + 1; V5V3; V^(V5+V^).
76 ANSWERS TO THE
Exercise 105.
1. 11; 10; f. ^^ (x = 7^^VTS,
2. 4; 15. * (2/ = A/i5.
c —
1 9n^
12. —;
34 m
^ ^^' "^'^^ f 2V«(V«+ v^)
iC =
a — b ^^ I "^ I) — a
; 54. 16. ^
I ^ ^ 2 (Va6l)
5. 2; ,V '^ ^^
6. 192; 3^. ^^ (^ = li,
7. 4.
8. 1. f _ Q^ — ^
18. ^^ln'
^^' 4m '^' ^i r m + y.^
 }
13. 15; 6. ^^ I Vm
14 ^(^ + ^) .
h
Vn
Exercise 106.
1. 0.7781; 1.8060; 1.1461; 0.9030; 1.0791; 1.1761; 1.9242,
2. 2.5353; 1.2040; 2.3343; 1.4313; 1.6532; 1.5562.
3. 1.9542; 2.3222; 3.5562; 3.0491; 3.2252.
ELEMENTS OF ALGEBRA. 77
Exercise 107.
1. 4; 1; 2; 2; 1; 1; 3; 7; 0.
2. 0.8281^ 2.8281; 1.8281; T.8281; 3.8281; 6.8281.
3. "Six;" "one;" "four."
4. "Fifth;" "first."
5. T.2552; 1.3522; 0.0212; 0.5741; 1.0212; 0.7993;
2.0970; 2.6232.
6. 3.7481; 1.1070; 1.1582; 0.0970; 1.0970; 2.6990;
5.4983.
7. T.4804; 0.7323; 5.7781; 3.3222; 0.5441; 4.5441;
0.6511.
Exercise 108.
1. 1.2040; 2.0970; 2.5350; 1.8060; 3.3397; 1.8060;
1.9084; 1.8572; 6.3588; 1.6110.
2. 2.5353; 2.5562; 4.1070; T5.2620 ; 3.5810; 5.2569.
Exercise 109.
1. 0.3980; 1.7781; 0.1461; T.7781 ; 0..5441 ; 5.6320;
T.3980; 2.4950; T.5441.
2. 0.6690; 4.1373; 0.0970; 3.7781; 1.5899; 4.1040;
0.6990 ; 0.4559.
Exercise 110.
1. 0.1690; 0.1590; 0.0602; T.9466 ; 0.7952; 0.6476;
T.9964 ; T.8950.
2. 0.8063 ; 0.3523 ; 1.2519 ; 0.0691 ; 0..3093 ; 0.5456.
3. T.6360; 1.6507; 0.2605; 0.6851; T.9962.
78 ANSWERS TO THE
Exercise 111.
1. 1.8451; 2.0086; 2.3032; 2.9996; T.8525; 0.5563;
S.8971 ; 0.5065.
2. 3.4914; 2.9926; 5.4771; 1.0034; 4.4692; 3.4794. .
3. 4.5142; 1.2638; T.5876; T.6235; 0.7672.
Exercise 112.
1. 2940 ; .0289 ; 63900 ; 3.151 ; 1.57.
2. .00455 ; 30.94 ; 33138 ; .03333 ; 4.566.
3. 7.586 ; 50.56 ; 633420 ; .001301 ; 539375.
Exercise 113.
1. 29.77 ; 4.75 ; 2.814 ; 32464.
2. 2.664 ; .368 ; .0769. 12. 9 ; 20 ; 121 ; 30 ; 56.
3. 373.6; .4847; .6186. 13. 2; 30; G; 14; 1.
4. 1.683; 2756; .482. 14. T.75729.
5. 9745; 3.264; 3.637; .4276; .2163.
6. 1.53+ ; 9.58+; 6.44; 5.59+.
7. 2.56+; 1.2+; 2. 15. 2; 3; 5; 7; 10.
8. 4.56+ ; 3.46+ ; 3. 16.  2 ;  3 ;  5 ;  6 ; ^.
log n log n — b log m log n
9. 3.3+
log m ' a log m ' a log m \ b log c
'09+, \ X = 1.177f
709+ ; '1?/=: .677+
^^ a::::.2.709+, 1 ^  1.177+, 17. g ; 3 ; 4 ; 6 ;  1.
\y = 1.
18. 4; 3; 3; 6.
_ 4 log m
"^"To^' (x = 3, 19. 2;;i;4.
\ _ ^Qg^ . h = ^ 20. y; t;2; 1;
^~ log 7^' 1.
ELEMENTS OF ALGEBRA. 79
21. 4; 16; 8; 64; 512; 32; 256; 1024; (2048)\
22. .00001; 5^3; 243.
23. 3; i; .3; V; V> and t?^; 4, and 4.
Exercise 114.
1. ± 3; ± 2\/A 8. ± Vm + n ; ± c.
a. ±2J; ±Vi. 9. ±\/^i ±V^f^'
3. ±8; ±5.5. 1^^
10. ± V t; ±4.
5. ±.3; ±tV2. ''»gl
6. ±2; ±^. "• ±"i±v!! *• *^
7. ±V30; ±}V77. 12. ±\/*(6»c2«;; ± J.
" c
Exercise 115.
1. 12,2; 8, 10; 16, 2. 7. ff,  y ; 0, 4.
2. 17,4; 4,13. 8. 2,4,5; 1,1,^.
3. 3, i; i, 3. 9. 2, V^; a, ^a.
4. 107,106; §,i. 10 '',^; 4,1.
5. ^,2; 2a, 8.
, o 11 1.^5 ±5, ±3\/2.
6. 0,3; T^, J. 'a 6'
Exercise 116.
1. 8, 15; 7, 6; \,\. 5. ?, i; V, ^3^
2. 23,1; 6, J/; 1,4. 6. V, §; A.?
3. 2,V; 3,1; 3,1. 7. 1,3; i,J; ^^J.
4. 11, V ; h 4; f A. 8. V, ^\ ^i f
80 ANSWERS TO THE
15. i(5±V22:6); 8,^.
16. 11, 2; 7, 2.
17. 5, 4f.
18. 9,H; 3,3§§.
19. 13, ±VM; 12,2.
Exercise 117.
5. 71)— p\ ± ^2n — ^/n.
/ — ; ha 1
2. ±yah\ , — T 6. a ± ; ^, — a.
a a
^ . "^h b 2b ^ . a
3. 1,  — ; , Q, Za, —a, a, —'
a — b a a 2
a^ __2y 8a a 5 =b V25 — 4:m^
a(a + ^) a (a — b)
9.
3,t; i,6§.
10.
1,14; ±9.
11.
13, §; 9,ff
12.
3, 8.7; 6, e.
13.
2, ^Y; 8± V601.
14.
12,2; 3,i.
1.
, « b
7. (a^± ^»)2;
a + b
9. (a + ^)^  {a  by; 0, ^ 4   c  1.
Til
10 ^ + ^ <^^ . a ± Va^ + 6'^
a — b' a + b' Sa^b^
a f ^b^  4 c'' \ l — 8a± 2aVa
^^' 2 V "^ 2«^ + ^ ; ' (a  ly
13. ± 1, ± ^ V^3; ± V2, ±^V6.
ELEMENTS OF ALGEBRA. 81
Exercise 118.
1. 9H,11; 2,4. 4. 2, i; 3,2.
2. 7,7; i,?^ 5. i,i; 3, f.
3. 4, 3S; 2,8. 6. 4a, ^ a; 2,22.
8. (1±V^); ^,a,^(5±Vr4); ±V;^1.
a \ d ac + bm^
»• 1' — d' "' ^ —
10. — , ^; m — 2 a, ^ m \ a.
n m,
11. a ± (6 — c). 16. 2 rt + 6 *, a — 8 ft.
2 * w — 2n' m ■\ 2n'
~^^\ 1 ± '
a + ft ± a/2 a^ + 2 ft*
13. ± V2 a  a^ 1 ± A/T^=^;  (2 ± Va^ + 4)
a
14.
4,/ _ 5ft a 2ft 2 /J7
15. ^ . , » . ' 18. ±^V3.
6aft 3aft 3a
Exercise 119.
2. 8, 9. 3. 15, 12. 4. 6, 21. 5. 8, 6.
7.
,»iO±\/^) ^1^
82 ANSWERS TO THE
9. Greater, _«y»L^; less, _^V" _ . g. g.
10. 18. 13. 87. 14. 53, 35.
11. Eate, 30 miles an hour; 7. 12. Kate, 6 miles an hour.
2m — n ±. ^/ii^ + 4 m?
16. La]'ger pipe, ^^ minutes ; smaller,
2m^r n± V^ + 4 m^ • ^ t • oo •
minutes. Larger pipe, 88 mm
utes; smaller, 154 minutes.
17. B, 6; A, 10. 18. 9 miles an hour.
19. miles an hour. 4 miles an hour.
14: am „
a± V^^ + ^
20. Length, ^— ^^^^'
weight, /, lbs. Length, 8 feet;
^ A arn, „ .,,.,,
— a±\/ \a^ weight, 4^ lbs.
21. <^VnTmVc Vn(VnTVc) ^^^^^^^.
^Jn^ \/c am
Vc(Vn^:Vc) ^^^^^^^ ^g ^^yg . j^ ^1 25 ; B, $1 ,
a — m
22. 10 (— 5 ± Vfn + 25) dollars.
23. 10(5 ± V25 — m) dollars.
Exercise 120.
1. ±2, fcVlO; 1,2; 2, 1, 1± ^=3, ^(1±a/^).
2. ~3; ± a, ±5. 4. 16, ifffi; 8, W
3. ±4, ± i; 9, (41)1 5. ± 8, ±(11)^; 9, ^/i681.
ELEMENTS OF ALGEBKA. t
6. 3», (28)1; (J)«, 1. 9 i JL .>_,., ,
tt 4 a * ^ ^
'■ '«.li 27. , 1
8. jm;«)u. 1,(196)". Venn (j^
10. ±^^/2^, ± \^^; ± vS, ± v'S.
Query. Wliy, iu 10, the ± sign ?
Exercise 121.
1. 2,  3, 1 ^ 3V5 . 9. i(3 ± V5), i(9 ± ^f^^\
2 5± V37 5± V7
2. 1, 1±2V15. "' 3 ' 3 '
3. ± 4 V6; 18^, 5. ^^' ' ^' _
 3 ± V33
4. 3, i, 4 C5 ± V1329). 13. 1^ , 1,2.
5. 5,6, Kl± V377). 14 1, 2, 5, 8.
® ' ^' 3 16. 5,  1, 2 ± V5.
7. 1,3±2V2. 3±a/29 3± \/I^
— ^'^' 4 ' 4
8. 4, 1,^^^. 18. 2^e^, ^rs.
10. 3, —1.
19. ^^^, 1± V2; 3,5, , 2 (13 ± V313).
Exercise 122.
1. 10,2; 7, 14. 4. 14, 2.48; V
2. 9, 12. 5. 4, 21; 5, \.
3. i, V; 8. I «• ^' 4' ^5 12, 4.
84 ANSWERS TO THE
7 __ 9 V2; 3, ^. Query. Is ^ a true value for x in the
second equation ? Why ?
8. 0, ± V3. 12 i^ ^ 4 B4
2m^/n ^ ^
9. 2^T. §; ± ^^ _^ ^ • 13. ± (^; 9, V
10. 0,4/7; ±Vl±h^2. 14. 1, 4; 9, 4, ~ ^ ^ " ^ '
11. ±*'(^')'; 25, V 16, 0, 2,3; ± ^^ V3.
_ 7±Vl3 lq=V^^ ,, , o^o./^
.u.
2" ' 2 ) ^'^j ^j ^ r
^ V t ,
Exercise 123.
1.
8,9; W, 49.
2.
. J;;,
mn
m — n^
3.
0, ^; a, a^ 15, 3.
2 b, y'a'.
5. a;2_4a.^21; Gcc^ + Sx^G; a;22a; = 15; x' = Z\
6. a;2 + 5a: = 0; x''Ux = —29', x^ — 2 x = 1.
7. ic'^ — (1 + m^) X = m {1 — rn^) ; mnx^ — (m^ — ?i^) a;
= m w ; (a — Z*) ic^ — 4 tt 6 ic = (ct — ^) (a + hf.
8. ic^ + 2 m a; = 8 /^ — m ; 4 ic'^ — 4 a ic = Z» — a""^;
Vox^ — '6\/'(ix^h — a.
Exercise 124.
1. Imaginary ; real and equal ; imaginary.
2. Eeal and rational, and different; imaginary.
3. Real and rational, and different; real and surds, and
different.
KLEMKNTS OF ALGEBRA 85
4. Real and equal ; imaginary.
5. Imaginary ; real and different, and surds.
6. Real and equal; real and rational, and different; real
and rational, and different.
7. Real and rational, and different, real and rational, and
different ; real and surds, and different.
8. Real and rational, and different; real and surds, and
different.
9. Im;i<^nnary; real and rational, and different.
Exercise 125.
1. ±8, ±(ll)i; h i; h 4".
2. 1^, T^; 1, ih' 3. I 450; 4, f
^"/ l ± \/l3 ^/ l ± V 7 J ± Ks ± 2 \/4 g + 1
V 2 ' ^ 2 ' ' 2
5. ± 3, — ; 5, 4,
6. ± h h h 7. 4, 1, 3, 2.
9. ± 2 \/2, ± \ a/=^, ± i V^185 ± 29 Vil.
„ n 3±4V3 , , "• 4m», to'; 8,4.
X2. 0, ^— , 1, 4. ^^ ^_ J
13. 4, 1; J, y. 16. 0, 1, 3.
86 ANSWERS TO THE
17. a±— ^=; 4, 3^3.
18. (i.) a/^^ — 4 ^ a surd ; (ii.) ^^ _ 4 ^ positive;
(iii.) ^2 _ 4^ negative; (iv.) yl^4j5 = 0;
(v.) J5 positive; (vi.) B negative;
(yii.) B = .
j^ .'±V4j^+a ,^g
Exercise 126.
12/ = 5, 4. ■ j2/ = 9,
2.
r.x = S, 1, J a; = 7, 5,
l2/=i,2. ■1j, = 5,7.
.^ = 6,4, 11 \x = %5,
'• ly = 3,7. ly = 5,9.
p= 6,^^,41, p = 7.4,1,
*J3, = 3i,4. 1^ = 7.8,5.
. rx = 18, 12J, <x = ll, 8,
=iy = 3,2i. "•1y = 8,ll
^•iy = 6,3. "•1^ = 1.
p = 4,3, ^^ . = 5,4,
y = 3, 4. l.y = 4, 5.
(x = 8,17§, ig fa; = ll,7,
° l2, = 6,13i. l3/ = 7.ll.
Query. In 14, has x and y two values? Why.'
1.
ELEMENTS OF ALGEBRA. 87
Exercise 127.
(x = ±5,±7y a: = 3, 3, 2,2,
U=±7, ±5. • ^2/ = 2, 2,3,3.
(X= i,
7, 4, f X = 5, — 5, 1, — , 1,
7. * j y = 1,  1, 5, — o,
p = 3,2, (x=±5, ±3,
^ U = 2,3. ''• y=±3,±5.
J a: = 5, 5, 3,7, j x = 10, 4.
*• 1^ = 3,7,5,5. • 1//4,10.
ra; = 6, 4, ^ ,3.= ± 2, ±3,
' (y = 4,6. • y=± 3, ±2.
p = 6,6,5,5, (0^ = 5,3,
^ y = 5,5,6,6. ^^ (y = 3,5.
Exercise 128.
r x = ± 2, ± 1,
• ly=±l, ±2.
rx=±2,
<x =
1.
{"
±3,
±2.
2.
!;:
±6,
± 2.
3.
i;:
±3.
4.
\l'
:±2,
: ±3.
5.
■■±h
■■±h
±i.
±3
7.
± 7, ± >v/3,_
±2, TSV3.
( ar = ± 3, ± 36,
• j y =: ± 5, :f ¥•
r X = ± 8,
1 y = =F 5,
^^^x=±8,±3.
88
ANSWERS TO THE
1. r
2.
3.
5. r
(y
6. f
7. f
3,5,
o, 3.
6,3,
3, 6.
8,7,
7,8.
10, 12,
12, 10.
10.
11.
12.
13.
Exercise 129.
14.]^
\y
15.
(2/ =
y
16. <
5, 3,l±iV88,
3, 5, 1 T iV88.
3,2,
2,3.
, o 13 ±V377
4, ^, ■ ^^^ ,
{y =
± 7, ± 5,
± 5, ± 7.
a:
2,
17.
3.
y
20,
15.
0, 5, 1,
18. .
(0!
0,1,1.
19. .
P
±10,
liS/
±4.
20. .
^Vf
21. <
a;
T 1, T 2 a/=3,
2/
± i, ± V
1.
0,4,
0,5.
22. .
[y
0,1,
0,2f.
23. .
y
y = 2, 4,
= 4,3,
= 3,4,
6
13:Fi
/377
6
7± V
295
2
7:f V
295
2,
2J,
4,
4.
6,
2,
2,
6.
2,
h
2,
16.
5 IF V15
2
5± Vi5
±iv/l8±2Vli
i(l±V15).
1,2,
3,1.
ELEMENTS OF ALGEBRA. 89
26
27
f 1 \ ab± V(« + 1) (fl^  1 ) (b + 1) (6  1)
^*' j «&  1 ± V(a + 1) (a  1) (b \l)(b 1)
1^^" ab
rx=±9, ±3, (:r = 2, i,
*^ iy=:±3, ±9. • 1^ = 3,24.
( X = ± 5, ± 2, ± 2 v^, ± 6 V^,
1 2/ = ± 2, ± 5, ± 5 V 1, ± 2 VH.
(x = ±l, ±3, jx = 3, 2, 3± v^,
(y= ±3,±1. • 1y = 2,3,3T V3.
„ ( « = ± 2, ± 3, ^ a: = 7, 1, 4 ± 2 V7,
''"• 1y=±3,±2. • ■iy:=:l,7,4:f 2V7.
30 ^
■[y = °a^V3),j(i.L).
rx = 2,4,3q:v'21, ( a: = 5, 4,
^^ y = 4,2, SiVai. ■<y = 4,5.
32. r=o'"!' •«>r==*^!'*l'
(y = 2,4. <y = ±4, ±9.
^^Tft' ■1y = l,9.
33.
35.
„=±?1±*!. „ 5a; = 62'S,l,
^ * a4 "• (y = 1,625.
x= ±4, itV^^eB, va6±V34lo),
90 ANSWERS TO THE
^x=± (ah), ^x = hh
^ =1,5. 50.
2/= ±^'
ly = ^,
46. .^2/ ^,5,,,
47. j^ = ^^' 52. 9 and 7.
Query. In 47, how many values has x and y ? Why ?
63. ± }^V^((^ + ^) and ± ^ V2 (tt  6). 11 and 7.
54. 36.
a + ^3a^ T 2 V2 (a"" + bj
55. __zi_I Z_ r — V — 2_y ^^^
Z
a + ^ 3 ^2 + 2 V2 (a^ + Z>) Q ^ 1 1 1 Q
—  — ^^ . 3 and 1, or 1 and 3
Z
56. — a and — 2 a, or 2 a and a.
57. 5 and 2, or — 2 and — 5, or ^ (3 ± V— 67) and
i ( 3 + V=^67).
58. 6.4, or 4.6. 59. 3, 15, and 20.
2 c" • , — a ± V4 c^ + a^
61. =:^= and . 8 and 4, or
. — a ± \/4: (§ + a^ 2
— 4 and 8.
62. f , or ^f. 60. 4 and 13.
63. Time, 7 years, or 6 years ; rate, .06, or .07.
64. Principal, $10400; rate, .05.
ELEMENTS OF ALGEBRA. 91
Exercise 130.
^ (x =11,8,5,2; x = 2,7,12,17, ....; a; = 8, 16, 24, 32, . ..
^' 1^ = 1,3,5,7; ^ = 7,21,35,49,....; i/ =5,8, 11, 14,....
^ a; = 42, 31, 20, 9; x = 215, 202, 189, 176, . . . . , 7.
^ ^y = 4, 9, 14, 19; y = 5, 14, 23, 32, ...., 149.
X = 8, 25, 42, 59, ,.. .; x = 7, 16, 25, 34, ....;
x = 4, 17, 30, 43, ....
y = 7, 22, 37, 52, ....; y = 10, 23, 36, 49, ....;
y = 2, 11,20,29,....
^x = 6;x = 9;x = 0. ^x = 10,
y = 3; 2/ = 3; y = 15. ' \y = 5.
IX = 3; x = 59; x = 476. ^ x = 4,
ly = '2;y= 1;3,= 19. " ^ = 24.
6.
r X = 11 ; X = 37.
jy=18;2/=13.
Exercise 131
11.
2.
59.
6.
\ and I
3.
13 and 1, or 4 and 8.
7.
131.
4.
72 and 70.
9.
Nine ways.
5.
( 5 foot rod ; 19, 12, and 5.
11.
No.
foot rod ; 4, 9, and 14.
10. Two ways in each. 39 and 6.
12. A gives B thirteen 50cent pieces, and B gives A forty
five 3^ cent pieces.
r Horses, 1,3, 5, 7,....
(Sheep, 8,23,38,53,....
92 • ANSWERS TO THE
rl6, 15, 14, 13, ....,at^25.
14. J 2, 5, 8,11,...., at $15.
157, 55, 53, 51, ...., at $10. Sixteen ways.
r 6, 13, 20, 27, 34, at 30 cents.
15. \ 45, 35, 25, 15, 5, at 45 cents.
1 24, 27, 30, 33, 36, at 80 cents. Five ways.
(74, 73, 72, ...., at 20 cents. .
16. ^4, 8, 12, .... , at 35 cents.
I 72, 69, 66, ...., at 40 cents. Twentyfour ways.
r23, 16, 9, 2, at $1.50.
17. \ 13, 16, 19, 22, at $1.90.
l 4, 8, 12, 16, at $1.20. Four ways.
18. Four ways. 21. 269.
19. 4 at $19, 8 at $7, and 8 at $6.
22. The first has 63 or 23, the second 37 or 77.
21, 23, 25, 27, poles 7 feet long.
23. ^ 18, 13, 8, 3, poles 10 feet long.
1, 4, 7, 10, poles 12 feet long. Four ways.
Exercise 132.
1.
a;>i; a^ < 4^
12 1"^^^'
2.
3.
x>^\ X <b
x>l.
1.
13. f>^^
9.
x = 6.
ly < ab.
10.
X = 4:.
15. 76 or 77.
11.
x = 5.
16. 60 cents.
ELEMENTS OF ALGEBRA. 93
Exercise 133.
8.i^2. a \ b 2ab m n ^ 1 .1
5. '^±y>t^. 3(l + a'^ + a^)>(l + a + a«)^
X — y X* — y^ ^ '
6. If«>/;, a» + 26«>3(^6^ :^ > :^; V2 4V7
_ v6 v3
> V3 + V5.
Exercise 134.
5. 4x^3x41 > 6. 18 j2^<^'
6. 42a'+;i2>3^^^ ' < x > 0.
7. 3 + A>a^ + 3:c«. 19 j^Jj^^^'
8. 3(a:y)^>3(ci4<^r; 1^ (x^ + y) > a;«  y»
10. lln'»>a2 + 8 6. ' \x >V
11. w^ < 54; „ c < 7i«^2^ ^ a; < _ ^,
12. 8>27; 15>8;5>3.
13. 3>4; 4m2 + 1>7i. 24. 2.
14. w > w ; 2 > — 4. 25. 6, or 7.
15. 2aj>m7i; ««>/. 26. 16, or 17.
16. x< 2.9; X <3i. 27. 19.
17. jc<; ar<i. 37. 13.
22. x> 5, ar<^; a;>i, x<§.
23. a; > §, X < i ; x > 7, x >  3.
28. 2, 1, 0, 1, 2, 3, 4; 8.
Vm ^ Vn
miles. 126 miles.
94 ANSWERS TO THE
Exercise 135.
1.
49.
5. SO a — 79 b.
9.
12.
2.
161, 245.
6. 1,0.
10.
19.
3.
16,9.
7. i.
11.
95.
4.
98, 243.6.
8. 103.
Exercise 136.
12.
2, 21, and 2^.
1.
779.
4. If (9.).
7.
76 a 4 57 h.
2.
 5569.
3.^
8.
8^.
3.
a^{4:a).
6. a7i(?z+l)
9.
13.
10. (1) 71 = 16, d = — l. (2) 7i = 7, d = 2 a.
11. 5^, 6^, 7t, etc. 13. 7500. 14. 5,1,3,7,11.
Exercise 137.
1 ^H'6A, ••••, ^TV 5. x''+lx,x^+22x,....,x.
7n 4 7i
2. 6.4, 5.6, ....,5.6. 6. , , .
3. 4 ?w — 5 ?z, 3 m — 4 w, — , 6 7i — 5 m.
4. _2l,3§, 4f, 5. 7. 101,7^.
1. 48; 384.
8. ^^W, Hi W, ••..
Exercise 138.
3 — ^t; — 2T^T
4. 128; 1.
6. x^»'
,288. 32^
^' 243'
« (§)'•
ELEMENTS OF ALGEBUA. 95
Exercise 139.
1. 2f
4. V(3+V3).
7. .'5.
520.
2. mh'
5. H3'"l).
8. 2.
3. mh
6. §(14'").
9. 6.
5, 30, 180.
Exercise 140.
1. 42; aH^', J V6; H 5. 40, 16, 6§.
2. ik; 6x2 — ox — 6. g _7^ ^, _, 1^ _^, ^^.
3. 20, 80. 8. Arithmetical, f .
4. i, i. 9. 2 and 8.
7. 6, 18, 54, 162, 486, 1458, 4374.
Exercise 141.
1. 4. 11. 1,9.
2. ^. 12. 20,5.
3. — ij't, — 5V' — ^> — ' iV« ^^f) —
4. 4, 2,$, I, f .... 13. 10,12,15.
5. If 14. i,i,i, 01^,1, J.
6 ''4,6. ^_,_^ 2a6
15. —^, Vab,
8. f , 5' A, h W, A. !«• 2550 yards.
9. 6, 12. 17. 3, 6, 12, 24.
10. 6, 2.
96 ANSWERS TO THE ELEMENTS OF ALGEBRA.
Exercise
142.
1.
54 6 : a.
10.
13.
2.
9:7.
11.
2, or 3.
3.
10 : 9.
12.
7: 2.
4.
(ly)(l + x):l+x\
13.
an — hm
5.
14.
m — n
h in — an
6.
5a:4; Ay : 6x\ Sy :
X.
m
7.
28 ic : By, n — 1 : 2a.
15.
4:7.
8.
7 : 8, 31 : 36, 41 : 48, 5
:6.
16.
"^ah.
9. a" — h^.a" \h' >a — h\a^h.
Exercise 143.
3. 4; 2; t33V210; 12; ^2^2^ 29. h '. a.
4. ^2 _ ^2 . 300 ^8 j^^ g^ ^^
5 M; 3r\; .8; 4^; 1^. 3,^ 4 . 1^
a?"
6. 3"; (a + ^)2; ^f,. 32. 17:7.
7. 15; i; 2^3^; 6^JZ». ^^ '"^ * ^•
8. 5/;2; 1. 34. 5: 4.
21. a = c' c?3, ^ = c^ (^i 35. V&xy.
23. a.= t^^a^^;a:3,orl. 37. ^ = ^' ■
^^ ' b n^
24.
( a; z!r i 4, i 6, ^^a + bV „ .
^ ^ / 39. £c = ^; — ^7 ); ir = 3,or— 1.
(y = ±6, ±4. W^y
27. A invested $3000 ; B, $3500.
38. Bate of slow train ; rate of fast train : : 1 : 2.
or THE ^ ^
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